Pivot point revision series 2011
Makesi mangi
PIVOT POINT REVISION MAGIC BULLETS 2011
CHAPTER 1
CARBON AND SOME OF ITS COMPOUNDS
Worked problems
Q (a) Draw the structure of;
(i) Diamond
(ii) Graphite
Answer
(i) Diamond
Each carbon is bonded to four others in a tetrahedral arrangement.
(ii) Graphite
Flat layer of graphite
Weak van der Waals force
between molecules
Flat layer of graphite with
hexagonal arrangement of
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carbon atoms. Each carbon
is bonded to three others.
(b) Using your knowledge of their structure explain the following properties of diamond and graphite.
(i)
Diamond is the hardest substance known while graphite is soft and greasy.
(ii)
Diamond is a poor conductor of electricity while graphite is a good conductor.
(iii)
Diamond has a density of 3.5g/cm3 while graphite has a density of 2.3g/cm3.
Answer
(i) Diamond has a giant atomic structure. To move any atom or particle on surface many strong covalent bonds have to be overcome.
On the other hand, for graphite the layers are only joined by Weak Van der Waal forces. The layers will easily slide over each
other and hence the soft and greasy nature.
(ii) Diamond is a poor conductor of electricity because it uses all of its four valency electrons to form covalent bonds. In graphite
each carbon only uses three of its valency electrons and the fourth electron remains free and delocalized.
(iii) Diamond has a higher density than graphite because all bonds in diamond are strong and the atoms are closely attracted. There
will be more atoms of carbon in a given volume of diamond than in graphite where we also have weak van der waal forces and
the atoms will be further apart causing a lower density.
(c)
Describe the appearance of
(i) Diamond
(ii) Graphite
(iii) Amorphous carbon
Answer
(i) Diamond is a colourless, transparent, sparkling solid with a definite crystalline shape.
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(ii) Graphite is a greyish-black, shiny crystalline solid.
(iii) Amorphous carbon is a black, dull solid. It is non-crystalline.
(d) Hot concentrated sulphuric acid or concentrated nitric acid react with
carbon.
(i) Write equations of reactions
(ii) What are the observations
(iii) Hydrochloric acid does not react with carbon. Explain this observation.
Answer
(i) 2H2SO4 (l) + C(s)
or 4HNO3 (l) + C (s)
CO2 (g) + 2H2O(l) + 2SO2 (g)
CO2 (g) + 2H2O (l) + 4NO2 (g)
(ii) In the case of reaction with Sulphuric acid effervescence occurs. Gas produced has a pungent smell.
Nitric acid produces a reddish-brown gas with carbon.
(iii) Hydrochloric acid has no oxidising properties therefore it cannot react with a non-metal like carbon.
(e) State three uses of carbon dioxide.
Answer
- Carbon dioxide is used as a fire extinguisher
- It is used in the manufacture of soft drinks
- It is used to deep freeze food and to keep ice-cream cold.
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Q2(a) The arrangement below was used to prepare and collect carbon dioxide. Concentrated acid was added to marble until
the gas started forming at a suitable rate.
Carbon dioxide
Water trough
Limestone
or marble
and acid
(i) Write equation of reaction between marble chips and hydrochloric acid (aqueous).
(ii) What is the test for carbon dioxide?
(iii) What is the method of gas collection above called?
(iv) If the gas was required dry.
I.
What method of gas collection would you use?
II. What would be a suitable drying agent for gas?
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III. Draw a suitable arrangement for drying and collection.
IV. What else could be included before drying if the gas was required pure. Explain your answer.
Answer
(i) CaCO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)
(ii) Mixing gas with lime water produces a white precipitate
(iii) Upward delivery over water (also called downward displacement of water)
(iv) I. Downward delivery over air (upward displacement of air)
II. Concentrated sulphuric acid or Calcium chloride.
III.
cardboard
cover
CO2(g)
Concentrated
Sulphuric acid
IV. Bubble gas through water to remove HCl(g) fumes.
(b) Write equation of reaction between carbon dioxide and;
(i) Sodium hydroxide solution
(ii) Calcium hydroxide solution
(iii) Sodium carbonate solution
(iv) Calcium carbonate in presence of water.
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Answer
(i) 2NaOH(aq) + CO2(g)
Na2CO3(aq) + H2O(l)
(ii) Ca(OH)2(aq) + CO2(g)
CaCO3(s) + H2O(l)
(iii) Na2CO3(aq) + CO2(g) + H2O(l)
2NaHCO3(s)
(iv) CaCO3(s) + CO2(g) + H2O(l)
Ca(HCO3)2(aq)
(c) What is seen in each of the reactions in b(i) and (iv) above
Answer
(i) A clear solution seen
(ii) A white precipitate is seen
(iii) A clear solution formed at first. A white precipitate will form later as NaHCO3 is not very soluble.
(iv) The solid disappears to give a clear solution
(d) When carbon dioxide is bubbled through water the product turns blue litmus to red. Explain this observation.
Answer
Carbon dioxide is an acidic gas it dissolves in water to form a weakly acidic solution
CO2(g) + H2O(l)
H2CO3(aq) (Carbonic acid)
(e) Burning magnesium were lowered into a gas jar of
carbon dioxide.
(i)
What would you see? Explain.
(ii)
Write an equation for the reaction.
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Answer
(i)
Black specks of carbon and a grey white residue. The intense heat of burning magnesium decomposes carbon dioxide to
carbon (black specks and oxygen). The oxygen released keeps the magnesium burning to produce magnesium oxide (white
powder)
(ii)
Mg(s) + CO2(g)
C(s) + MgO(s)
(f)(i) Below is a diagram of a luminous flame from bunsen burner. Label diagram fully.
Burner tube
Collar
gas
jet.
NB/ Jet is just a pipe
Base
(ii) State at least two differences between a luminous and a non- luminous flame.
Answer
(i)
I.
Almost colourless mantle
II.
Bright yellow zone
III.
Dark Central zone
IV.
Blue Zone.
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(ii)
Luminous flame
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Non-Luminous flame
- Cool sooty flame
- Much hotter and cleaner flame
- Large part of flame is yellow
- Large part of flame is an
almost colourless zone
- Blue zone is very small
- Blue zone is much bigger
3(a)(i) Carbon monoxide burns in air. Write equation of reaction.
(ii) The reaction;
C2H2O4.5H2O + H2SO4(conc)
CO2(g) + CO(g) +CO(g) + 6H2O(l)
+ H2SO4(aq)
I. How is the carbon monoxide formed separated and collected from
this reaction.
II. State two properties of carbon monoxide.
Answer
(i)
2CO(g) + O2(g)
2CO2(g)
(ii) I.
Pass the gaseous products through sodium hydroxide solution and
collect Carbon monoxide over water.
II. Carbon monoxide is neutral to litmus paper and only slightly soluble
in water.
- Carbon monoxide is poisonous.
(b) In an experiment gas Y was passed over heated charcoal and product passed over heated Iron (III) Oxide. Gases were then passed
through lime water which formed a white precipitate. Gas left burned with a blue flame at a jet.
(i)
Identify gas Y
(ii)
Write the equation of reaction between
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I. Gas Y and charcoal
II. Iron and product from reaction with charcoal
(iii)
The charcoal and iron were in separate porcelain boats. What was seen in each case.
Answer
(i) Gas Y is Carbon dioxide
(ii) I. CO2(g) + C(s)
2CO(g)
II. Fe2O3 + 3CO(g)
2Fe(s) + CO2(g)
(iii) The charcoal disappeared as it formed gas and the brown iron(III) Oxide formed a shiny greyish solid.
(c) Below is a diagram of a coke fire. Study it and answer the questions that follow.
(d)
Chimney
Z
Air
Y
Air
X
(i) Write an equation for reaction in each of the regions X, Y and Z.
(ii) What is seen at Z?
(iii) Why is it not advisable to light a charcoal “jiko” in a poorly ventilated room.
Answer
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(i) C(s) + O2(g)
CO2 + C
2CO(g) + O2
CO2(g)
CO(g)
Makesi mangi
(X)
(Y)
2CO2(g) (Z)
(ii) A blue flame
(iii) Such a fire will produce carbon monoxide which is poisonous and may lead to death in a poorly ventilated room.
(e) Carbon can be used to reduce oxides of lead and copper
(i) Write the equations of reaction
(ii) What would be seen in each case.
Answer
(i)
(ii)
C(s) + 2CuO(s)
2Cu(s) + CO2(g)
2PbO(s) + C(s)
2Pb(s) + CO2(g)
The black copper oxide changes to a red brown shiny solid.
and the yellow solid changes to a grey shiny solid.
4. Below is a flow diagram of the stages for manufacturing sodium carbonate in Solvay process.
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Brine
Na HCO3
water
NH3
Chamber C
Product Y
water
Product Z
NH4Cl
Filtration
Steam to
provide energy
for reaction
NH4Cl
C
Calciner
product W
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CO2 + H2O
Limestone
Storage and
packing of
Na2CO3
Product X
(a) (i) State the products W, X, Y and Z.
(ii) State what happens in chambers A, B and C fully.
Answer
(i) W = CaCl2 , X = Carbon dioxide
Y = ammonia, Z = mixture of NaHCO3 and NH4Cl
(ii) In A, CaCO3 is decomposed by heating to produce CaO and CO2.
In B, ammonium chloride reacts with CaO to produce NH3 and CaCl2
In C, Brine is mixed with other reagents to produce NaHCO3 and NH4Cl. A lot of heat is produced and hence the need to circulate
cold water round the chamber.
(b)(i)
Explain what happens in the calciner using an equation.
(ii)
Name the compounds recycled.
Answer
(i)
Sodium hydrogencarbonate is decomposed by heating.
2NaHCO3
(ii)
Na2CO3(s) + CO2(g) + H2O(l)
Ammonia, Carbon dioxide and water.
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(c)(i) What property of Sodium hydrogen carbonate enables its separation
from ammonium chloride?
(ii) What property of reaction in chamber C makes it necessary to circulate cold water round it?
(iii) Identify three exothermic and three endothermic reactions in the Solvay process.
Answer
(i) Sodium hydrogen carbonate has a low solubility therefore it precipitates.
(ii) Reaction is exothermic
(iii) Exothermic reactions
NH3(g) + H2O(l)
NH4OH(aq)
CaO(s) + H2O(l)
Ca(OH)2(aq)
2NH4OH(aq) + CO2(g)
(NH4)2 CO3(aq) + H2O(l)
Endothermic Reactions
CaCO3(s)
heat
CaO(s) + CO2(g)
2NH4Cl(aq) + Ca(OH)2(aq)
2NH3(g) + CaCl2(aq) + 2H2O(l)
2NaHCO3 heat Na2CO3(s) + CO2(g) + H2O(l)
(d) (i) Write the equation of reaction in chamber C.
(ii) Considering the reactants, circulated by-products and the final products.
I. Write the overall equation of reaction in the Solvay process.
II. Why is the reaction not done directly.
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Answer
(i)
NH3(g) + NaCl(aq) + CO2(g) + H2O
(ii) I. CaCO3(s) + 2NaCl
NH4Cl(aq) + NaHCO3(aq)
Na2CO3 + CaCl2
II. Insoluble Calcium carbonate cannot react with Sodium Chloride.
(f) From 1.50 tonnes of salt mixed with other reactants, 1 tonne of impure soda ash was produced.
(i) If the soda ash is 58% pure calculate the mass of pure, sodium carbonate.
(ii) What is the percentage conversion of sodium chloride to soda ash.
(iii) Give one reason why yield may be low.
Answer
(i)
Impure soda ash is 1.0 tonne
Pure soda ash =
58x1.00
100
= 0.58 tonne
(ii)
2NaCl
= 1 Na2 CO3
i.e. see d(ii) or other relevant equation.
Na2CO3
= 23x2+12+48 = 106
NaCl = 23+35.5 = 58.5
2NaCl = (58.5x2) = 117
Mass of Na2CO3 expected from 1.5 tonnes
= 106x1.5 tonnes
117
= 1.36 tonnes
But actual yield = 0.58 tonnes
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Therefore percentage yield
= 0.58 x 100% = 42.6%
1.36
(iii)
Not all the Sodium hydrogen carbonate remains as precipitate. Some dissolves.
5. The following is an outline of the Carbon cycle. Study it and answer the questions that follow.
Plants
Food
Process A
Breathing
Wood coal
and decay
Animals
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process E
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Carbondioxide
in
atmosphere
Process B
Breathing, Decay
Process C
Product X
Carbonic acid
Process D
Pressure and heat
Hard water
Shells
Animals
(a) State processes A, B, C , D and E
Answer
(a) A is photosynthesis
B is dissolution of CO2 in water
C is Heating
D is a reaction to form a hydrogen carbonate with carbonic acid.
CaCO3(s) + H2O(l) + CO2(g)
Ca(HCO3)2(aq)
E is burning of wood or coal.
(b)(i) Name product X
(ii) Explain how product X produces carbon dioxide
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Answer
(i)
Limestone, chalk or marble
(ii)
It is decomposed by heat
CaCO3(s)
Heat CaO(s) + CO2(g)
(c) State two ways in which breathing is related to process C.
Answer
- Both use Oxygen and release CO2
- Both processes produce energy.
(d) State six elements in coal
Answer
Carbon, hydrogen, Oxygen, nitrogen, Phosphorus and sulphur.
(e)
Use the correct terms to describe:
(i)
Two processes that remove carbon dioxide from the air
(ii)
Three processes that add carbon dioxide into the atmosphere.
Answer
(i)
Photosynthesis and dissolution of CO2 into the sea
(ii)
Burning of wood and coal
Heating of carbonate rocks to decompose
Breathing and decay
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PRACTICE QUESTIONS
1. a) The arrangement below was used to study the reducing action of gas X.
Copper II Oxide
Lead II Oxide
Gas X
Excess gas X
heat
heat
burning with
blue flame
Anhydrous
Combustion
Calcium chloride
Porcelain boat
Lime water
(i)
Lime water in conical flask formed a white precipitate. What is the identity of gas X.
(ii)
What is observed on each oxide in combustion tube
(iii)
Write equations of reactions between gas X and heated oxides in boats?
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(iv)
What is the purpose of anhydrous calcium chloride.
(v)
Name another gas that burns with a blue flame and write an equation for its combustion.
(b) Table below is a record of incomplete observations on heating carbonate and hydrogen carbonates in a test tube.
Carbonate
Observation
Magnesium carbonate
Zinc carbonate
Lead carbonate
Crackling sound, gas forms
white precipitate with
limewater. Yellow residue but
brown when hot.
Copper carbonate
(i)
Complete the table
(ii)
Write equation of reaction for the effect of heat on each of the three carbonates.
(c) The arrangement below is an incomplete table of tests on residue obtained on heating carbonates and hydrogen carbonates.
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Carbonate /
Residue in
Residue
Adding dilute
Hydrogen
crucible
added to
Hydrochloric
water and
acid to residue
carbonate
stirred and
tested with
litmus
Calcium
White
carbonate
residue
Sodium
Effervescence
carbonate
Lithium
White
Solution turns
carbonate
residue
red litmus
No effervescence
blue
Sodium
hydrogen
carbonate
Calcium
hydrogen
carbonate
(i)
Complete the table
(ii)
Write equation of reaction (if any) for each of carbonate or hydrogen carbonate.
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(iii)
Makesi mangi
What is the common name for
I.
Sodium carbonate
II.
Sodium hydrogen carbonate
III.
Calcium carbonate
IV.
Calcium oxide
V.
Calcium hydroxide solution
VI.
Calcium hydroxide
(iv) To carry out the heating of calcium carbonate, a strong open flame was used in the following arrangement.
Iron rod
Limestone
Heat
(Hot flame)
Cricible
Residue
Why do we need a strong open flame to heat calcium carbonate?
(d) State two uses of
(i)
Sodium carbonate
(ii)
Sodium hydrogen carbonate.
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(Q2) Below is a passage describing how Sodium chloride is obtained from Lake Magadi in the Rift Valley. Study it and answer
questions that follow.
Lake Magadi consists of massive deposits of trona (Na2CO3.NaHCO3.2H2O) a crystalline solid. This solid material has formed
as a result of the high rate of evaporation on this water mass with high salt content. Below the trona is a mother liquor of Sodium
chloride and Sodium carbonate with a density of 1.26x103kgm-3.
The less soluble salt crystallises out first. This knowledge is made use of in separation of dissolved solids; a process known as
fractional crystallisation. Sodium chloride and sodium carbonate have the same solubility at 36.1 0C but sodium chloride is more
soluble below this temperature while sodium carbonate is more soluble above this temperature.
During the process of obtaining sodium chloride mother liquor is made to flow into channels and then pumped into salt fields
where initial evaporation by solar energy occurs. These are the pre-concentration ponds where the liquor flows from one pond to
another by gravity. Precipitation of trona and sodium carbonate monohydrate occurs.
Eventually the concentration of common salt equals that of sodium carbonate and the two form 16% of the mother liquor. The
liquor is pumped into salt making ponds where further evaporation occurs.
During hot months of the year the daily temperatures of this liquor can be as high as 400C but generally below 36.10C. During
night temperatures average 200C. Because of the difference in solubility of salt and carbonate and the high variation of
temperatures between day and night, the two deposit at different times. Impure salt is obtained which is purified by washing with
saturated sodium chloride solution. The purified salt is then iodised to produce coarse salt. Some of the coarse salt will be treated
by adding red oxide (Iron (II) Oxide for cattle.
Coarse salt contains 99.5% NaCl, 0.03% NaF,
0.4% sodium carbonate and 0.0037% iodine.
(a) What is the mass of water in 1000kg of mother liquor with density of 1.26x103kgm-3?
(b) What does the precipitation of trona and sodium carbonate monohydrate in the pre-concentration ponds tell you about the relative
concentration of sodium carbonate compared to sodium chloride in the lake?
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(c) What is the mass of sodium chloride in 1000kg of mother liquor when concentration of salt equals concentration of sodium
carbonate?
(d) Which of the two is precipitated during day, sodium chloride or sodium carbonate?
(e) Write formula of;
(i)
Sodium carbonate monohydrate
(ii) Sodium hydrogen carbonate
(iii) Sodium fluoride
(iv) Iodine
(f) To make 1000kg of coarse salt how much
(i)
Sodium chloride sodium fluoride and sodium carbonate would be weighed?
(ii) Iodine would be weighed?
(g) State two other uses of sodium chloride?
(h) Sketch solubility curves for sodium chloride and sodium carbonate against temperature using the table given on the grids below;
Temperature (0C)
0
10
20
30
40
(i) Sodium carbonate
7.0
12.5
21.5
40.1
48
(ii) Sodium chloride
35.
35.8
36.0
36.2
36.5
Solubility in grams of
solute in 100g
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Q3. (a) (i) What is allotropy
(ii) Explain how carbon is able to make marks on paper and at the same time scratch marks on glass.
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(b) Coke is a form of carbon obtained when coal is heated strongly in absence of air (carbonized). Study the reaction scheme below
and answer the questions that follow.
O2, heat
Coke
gas Y
Silica 30000C Silicon carbide + carbon monoxide
sulphur
10000C
Liquid Z, Cl2, Catalyst
Metallic Oxides
Liquid M + S2Cl2(l)
10000C – 15000C)
quicklime
20000C
Metal
CaC2 + CO(g)
H2O
C2H2 + Solid X
(i)
Name gas Y, liquid Z, liquid M and solid X.
(ii)
Write a balanced equation of reaction between silica (a binary compound of silicon) and coke.
(iii) What method would you use to separate the two liquids M and disulphur dichloride
(iv) Write a balanced equation of reaction between calcium carbide and water.
(v)
What is quicklime?
(vi) What property of carbon is illustrated by its reaction with metallic oxides.
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(vii) Choose a relevant example of metallic oxide and write its reaction with carbon (coke)
(c) (i) Write three characteristics of wood charcoal and three for graphite.
(ii) How would you illustrate that the two are allotropes of one another.?
4(a) Atomic number and relative atomic mass of carbon are 6 and 12
respectively
(i) Draw atomic structure of carbon
(ii) Name any covalently bonded compound of carbon and draw its electronic diagram.
(b)
How are the atoms
(i) Arranged in graphite?
(ii) Arranged in diamond?
(c) Use your knowledge of the structure of graphite and diamond to explain at least three of their physical properties.
(d) Diamond and graphite both have giant structure while iodine crystals has iodine molecules;
(i) Explain the statement
(ii) Show how the physical properties can be related to structure
(e) In which group and period is carbon found in periodic Table.
Q5. (a) Below is a diagram of arrangement of apparatus used to study products of a burning candle. Use it to answer the
questions that follow.
beaker
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To filter pump
Ice clods
Lime water
Filter tunnel
(i) Compare the flames when the filter pump is on and when it is off.
(ii) What was observed in the limewater?
(iii) What was observed in the U-tube.
(iv) What is the test for product X?
(v) Explain your observation in (ii) and (iii) above?
(vi) What do your observations tell you about candle wax?
(b) 224cm3 of methane, CH4 and 112cm3 of ethane, C2H6 mixture were burnt in oxygen.
All volumes were measured at s.t.p.
(i) Write equation of reactions
(ii) Calculate the volume of oxygen required to burn the mixture of gases at s.t.p. conditions.
(H = 1, C = 12, O = 16, 1mole of gas at s.t.p. is 22.4cm3)
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CHAPTER 2
MOLE CONCEPT:
Formulae and chemical equations, gas laws
Solved problems;
1(a) Explain each of the following in terms of particulate nature of matter.
(i) When a teaspoonful of sugar is stirred in water in a glass, it dissolves and disappears,
(ii)When ice is heated, it melts into a liquid and then eventually boils into a gas
(iii) Lead bromide does not conduct electricity unless it is molten.
(iv)
When a crystal of potassium permanganate is put in water in a beaker and left undisturbed for about 1 hour, the purple
colour spreads throughout the liquid.
(v) When heated, mercury oxide produces two substances with very different properties.
(vi)
Very small particles of carbon suspended on the surface of water display a Zig-zag motion.
Answer
1(a) (i) The water molecules pull the sugar molecules apart and come, between them and separate them to form a solution.
(ii) Ice is made up of water particles (molecules) held in fixed positions.
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When heated the molecules gain Kinetic energy and soon have enough
energy to overcome the attractive forces. When
heated further, more energy will be absorbed until the attractive forces in the liquid will be completely overcome, then gas
begins to form steadily.
(iii) The particles in lead bromide solid are held in rigid fixed positions and hence cannot conduct. In molten state the particles
(ions) are made free to move and it conducts.
(iv) Water molecules pull the particles of potassium permanganate and go between them to dissolve it. Once separated they
diffuse within the liquid.
(v) Particles of mercury oxide decompose to give particles of mercury and oxygen. Mercury and oxygen have properties which
are very different
(vi) Water particles move and hit the suspended carbon particle causing it to move. Its direction of motion changes as it
experiences different collisions each time.
(b) The volume of a given mass of gas is 300cm3 at 400c and 650 mmHg pressure.
(i) Pressure is increased to 1300 mmHg pressure at the same temperature
(ii)Both pressure and absolute temperature are doubled.
Answer
b(i) According to Boyle’s law, pressure is inversely proportional to volume at fixed temperature.
P  1/V , constant temperature
Therefore PV = constant
= 650x300 = 1300x V1 Where V1 = vol. at pressure of 1300mmHg.
Therefore volume = 150cm3
(ii) Temperature = (80+273)K = 3530K
but initial temperature = (40+273)0K
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= 3130K
Charles’ law:
Volume of a gas is proportional to temperature at constant pressure
i.e V  T at constant pressure
Therefore V/T = constant.
300 = V2
313
Where V2 = volume at 800c.
353
V2 = 300x353
313
= 338.3cm3
(iii)
The volume will remain unchanged at 300cm3. Doubling the absolute temperature has an exactly equal and opposite effect to
that of doubling the pressure.
(c) (i) Define the term diffusion.
(ii) State Grahams law of diffusion.
(iii) 100cm3 of oxygen diffuses through a porous plug in 60 seconds.
Calculate the rate of diffusion of oxygen.
(iv)
Hydrogen is 16 times less dense than oxygen;
I. Which one will diffuse faster, oxygen or hydrogen.
II. How long will it take hydrogen to diffuse through the same plug.
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Answer
c.(i) The spreading of a gas from a region of low concentration to one of high
concentration.
(ii) Graham’s Law states that the rate of diffusion of a gas is inversely proportional
to the square root of its density or relative molecular mass under the same
conditions of temperature and pressure.
(iii) Rate of diffusion = Volume of gas diffused
Time taken
100
=
/60 cm3/sec
= 12/3cm3/sec
(iv) I. Hydrogen.
II. Volume diffused same therefore in place of relative rates we can use relative
times. But time is inverse to rate. Let time for H2to diffuse be t H2
Therefore to2 = Mo2 =  do2 = 16 = 4
tH2
MH2
dH2
therefore tH2 = to2 = 60 = 15sec
4
4
Q2(a) What name is given to the instrument used for determining relative atomic masses?
Answer
(a) Mass spectrometer
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(b) The balance below illustrates an idea on relative atomic masses. The number of carbon atoms that balance the mass of a
magnesium atom are shown.
Mg
C
C
Each C=12
(i) What is the relative mass of a magnesium atom?
(ii) 24 atoms of hydrogen would balance the two carbon atoms. What is the mass of an atom of hydrogen?
Answer
(b) (i) R.A.M of Mg = (12x2) = 24
(ii) Let the mass of a hydrogen atom be h,
therefore
24h = (12x2)
h = (24/24) = 1
Therefore R.A.M. of H = 1
(c) Define (i) Relative atomic mass
(ii) A mole
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Pivot point revision series 2011
Makesi mangi
Answer
c(i) Relative atomic mass is the mass of an atom on a scale where one unit is
1
/12th of the mass of a C-12 atom.
(ii) A mole is a quantity obtained by expressing relative mass in grams.
(d) Determine the relative atomic mass of:(i) Chlorine, given that chlorine –35 and chlorine –37 exist in the ratio
754:246
(ii) Silicon, given that silicon –28, silicon –29 and silicon –30 are 92.2%, 4.7% and 3.1% abundant.
Answer
(d) (i) Total ratio parts = 754+246 = 1000
R.A.M. of chlorine = (754 x 35 + 246 x 37)
1000
= (35492/1000)
= 35.5
(ii) Assume that we have 100 atoms
R.A.M. of silicon = (28x92.2+29x4.7+30x3.1)
100
= (2810.9/100)
= 28.1
(e) What are the differences among the terms relative atomic mass, relative molecular mass and relative formula mass?
Answer
All refer to relative masses but some particles are only atoms so we determine their relative atomic masses. Some particles are ionic
compounds so they are represented by formula units. In this case the particle represented is only a ratio formula and the relative mass
is best referred to as relative formula mass. Some particles are molecules and so their relative mass is for the molecules. However all
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Pivot point revision series 2011
Makesi mangi
the relative masses are referring to some formula and so they can all be refferred to appropriately by using the term relative formula
mass.
(f) Determine the relative formula mass or relative molecular mass of;
(i)
FeCl3
(ii)
CuCl2
(iii)
H2O2
(iv)
Al2(SO4)3
(v)
Na2CO3. 10H2O
(vi)
Fe2(S2O3)3
(Fe = 56, Cu = 63, Cl = 35.5, H = 1.0, Na = 23.0, Al = 27)
Answer
(e) (i) FeCl3 = (56+35.5x3) (ii)
CuCl2
= 63+35.5x2
= 162.5
(iii)
= 134
H2O2 = 1 x 2 +16 x 2 (iv)
Al2(SO4)3 = 27 x 2 + (32 + 64) x3
= 34
(v)
= 342
Na2CO3.10H2O = (23 x 2 + 12 + 16 x 3) + (2+16)x10
= 286
(vi)
Fe2(S2O3)3 = 56 + (32 x 2 + 48) x 3
= 56 + 112 x 3 = 392
Q3(a) Calculate the number of moles of;
(i)
Sulphur in 200 grams of its pure solid
(ii)
Oxygen atoms in its 8.0 grams
(iii) Oxygen molecules in 16.0 grams of the gas
(iv) Oxygen molecules in 1.5x1023 atoms of the substance
(v)
Copper in 1.5x1023 particles of its atoms.
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Makesi mangi
(S=32, O=16, Cu=63; L=6.0x1023mol-1)
Answer
(i)
Mole of Sulphur
= Mass
=
Relative mass
20 = 0.625 mole
32
= (8/16) = 0.5mole
(ii)
Mole of Oxygen
(iii)
Mole of oxygen molecules = (no. of particles)
2xL
=
1.5x1023
2x6.0x1023
=
(iv)
Mole of copper
0.125mole
= no. of atoms
L
= 1.5x1023
6.0x1023
= 0.25mole
(b) Calculate the number of particles of;
(i) Sodium atoms in its 2.3grams
(ii) Sulphur atoms in its 0.2 mole
(iii) Oxygen molecules in 8.0 grams of the gas.
(iv) Oxygen atoms in 0.5 mole of the gas
(v) Magnesium atoms in its 0.12 grams of the metal.
(L=6.0x1023mol-1, Na=23.0, S=32.0, Mg= 24.0)
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Answer
b(i) 23g Na = 6.0x1023atoms
Therefore 2.3g = (2.3x6.0x1023)
23
= 6.0x1022 atoms of sodium
(ii)
Atoms of sulphur = moles x L
= (0.2x6.0x1023)
= 1.2x1022 atoms
(iii) Oxygen molecules
= moles of atoms x L
2
= (8.0 x ½ x 6.0x1023)
16
= 1.5x1023molecules
(iv) Atoms of oxygen = moles x L
= (0.5x6.0x1023)
= 3.0x1023 atoms
(v) No. of atoms of Mg = moles of atoms x L
=( 0.12 x 6.0x1023)
24.0
= 3.0x1021 atoms of magnesium
(c) Calculate the mass of;
(i) 6.0x1023 molecules of oxygen
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Pivot point revision series 2011
Makesi mangi
(ii) 6.0x1023 atoms of oxygen
(iii) 6.0x1023 molecules of phosphorus (P4)
(iv) 0.25 moles of carbon dioxide
(v) 2 mole of carbon dioxide
(O=16, C=12, P=31, L=6.0x1023 mol-1)
Answer
c(i) 1mole = 6.0x1023particles (O2)
6.0x1023 molecules = (16x2) grams
= 32 grams
(ii) 6.0x1023 atoms of oxygen = 1 mole of atoms
= 16 grams
(iii) Moles of P4 molecules = no. of molecules
L
= (6.0x1023) = 1
6.0x1023
Therefore, mass = (31x4)g
= 124grams
(iv) CO2 = 12 + 16 x 2 = 4
But mass = moles x relative mass
Therefore mass = (0.25 x 44) = 11g
But CO2 = 12 + 16 x 2 = 44
v) Mass
= Mole x relative mass
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Makesi mangi
=2x4
= 88 grams
(d) Below is a table of particles, relative mass and mass of 1 mole, 6.0x1023 particles. Study it.
Particles
Relative mass
Mass of 1mole
Carbon, C
12.0
12.0 grams
Sodium, Na
23.0
I
Oxygen, O
16.0
II
Oxygen, O2
III
32.0 grams
(i)
Complete table
(ii)
How many particles are found in 1 mole of substance?
Answer
(i)
I, 23.0 grams. II. 16.0 grams III. 32
(ii)
L, 6.0x1023 particles
(e) Calculate the percentage of the element stated below in the hydrated compound, Iron II sulphate, FeSO 4.7H2O
(i)
Iron
(ii)
Sulphur
(iii)
Oxygen
(Fe = 56.0, S = 32.0, O = 16.0, H = 1.0)
Answer(d)
(i)
FeSO4.7H2O = 56+32+16x4+18x7
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Pivot point revision series 2011
Makesi mangi
= 278
Therefore percentage, iron
= 56 x 100%
= 278
= 20.1%
(ii)
Percentage of sulphur
= 32 x 100%
278
= 11.5%
(iii)
Oxygen
= 16x4+7x16 = 168
Therefore percentage, Oxygen
= 168 x 100%
278
= 60.4%
Q4(a) Calculate the loss in mass that occurs when 200 g of calcium carbonate is heated t o a constant mass
Answer
(a)
CaCO3 = 40+12+16x3 = 100
CaCO3(l)
1mole
But 1 mole CaCO3
heat CaO(s) + CO2(g)
1mole + 1mole
= 100g
2 mole CaCO3 = 200g
Now CaO
= 40+16 = 56
Therefore 200g CaCO3 will produce (56x2) = 112g CaO
(b) Calculate the mass of ammonium chloride that will just react completely with 20.0g of calcium hydroxide.
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Makesi mangi
(Ca=40.0, O=16.0, H = 1.00, N=14.0, Cl = 35.5)
Answer
Ca(OH)2(s) + 2NH4Cl 2NH3(g) + CaCl2(aq) + 2H2O(l)
Ca(OH)2 = 40+(16+1)x2 = 40+34 = 74.0
NH4Cl = 14+4x1+35.5 = 53.5
Therefore; Mass ratio is 74:53.5x2 = 74:117
Therefore 20g of Ca(OH)2 = (117.0 x 20)
74
= 31.6g NH4Cl
(c) In (b) calculate the moles of ammonia produced.
Answer
Moles of Ca(OH)2
= (20/74) = 0.270mole
But mole ratio is 1:2
Therefore, mole of NH3 = (0.270x2)
= 0.54mole
(d) 10 grams of sodium hydrogen carbonate was heated until a constant mass produced.
(i) Write an equation for the reaction
(ii) Calculate the mass of residue formed
(iii) How many moles of carbon dioxide was produced
Answer
(i)
Reaction
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Pivot point revision series 2011
2NaHCO3(s)
(ii)
heat
Makesi mangi
Na2CO3(s) + CO2(g) + H2O(l)
NaHCO3 = 23.0+1.00+12.0+16.0x3
= 84
Na2CO3 = 23.0x2+12+48.0
= 106
Therefore mass ratio is
84x2:106
= 168 : 106
Therefore mass of Na2CO3 = 106 x 10
168 1
= 6.31g
(iii)
Mole ratio is 2:1
(i.e. mole NaHCO3 : CO2)
Therefore mole CO2 = ½ x (10/84)
= 0.119 mole
1(a) Determine the empirical formula for the following molecules.
(i)
Glucose, C6H12O6
(ii)
Ethene, C2H4
(iii) Water, H2O
(iv) Hydrogen peroxide, H2O2
Answer
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(i) Ratio of atoms is 6:12:6 = 1:2:1
E.F. is CH2O
(ii) Ratio is 2:4 = 1:2
Therefore E.F is CH2
(iii) Ratio is 2:1
Therefore, E.F. is H2O
(iv)
Ratio is 2:2 = 1:1
Therefore, E.F is HO.
(b) Given the following data, determine the empirical formula in each case:
(i)
Compound contains 8.85, of hydrogen, and 41.15g of nitrogen
(H = 1.00, N=14.0)
(ii)
An oxide of copper weighing 0.63g was reduced by hydrogen to form a brown residue of mass 0.508g (Cu=63.5; O=16.0)
(iii) A compound having 52.9% aluminium, 47.1% oxygen (Al=27.0, O=16.0)
(iv) A compound of 6.48g of magnesium combined with 2.1g of nitrogen
Answer
(i)
H
N
Mass
8.88g
41.15g
Mole
8.85
41.15
1
1.4
8.85
2.94
(8.85)
(2.94)
2.94
2.94
3.01
1.00
Ratio
=
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Pivot point revision series 2011
Makesi mangi
Therefore, E.F. is NH3
(ii)
Cu
O
0.636
Mass
0.508
0.128
0.508-
Mole
(0.508)
(0.128)
0.128g of
63.5
16
Oxygen
0.008
0.008
(0.008)
(0.008)
0.008
0.008
1
1
Ratio
Therefore, E.F. is CuO
(iii)
Al
O
Mass
52.9
47.1
Mole
(52.9/27)
(47.1/16)
1.96
2.94
(1.96)
(2.94)
Ratio
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Pivot point revision series 2011
1.96
1.96
1
1.5
2
3
Therefore, E.F. is
(b)(iv)
Mg
Al2O3
N
Mass 5.35g
2.1g
Mole (5.33/24)
(2.1/14)
0.222
0.15
Ratio (0.222/0.15)
(0.15/0.15)
1.48
1
2.96
2
Approx.
Makesi mangi
3
2
(nearest whole number)
Therefore, E.F. is Mg3 N2
(c) Hydrated crystals of a salt have a formula, Na2SO4.pH2O where p is a whole number. 1.61g of crystals were heated to
dryness and 0.71g of anhydrous salt produced. Find the value of p.
(Na = 23.0 , S = 32.0 , O = 16.0)
Answer
Na2SO4 = 23x2+32+16x4 = 46
32 +
64
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Makesi mangi
142
H2O = 1x2 + 16
= 18; Mass of water of crystallisation lost
1.61
0.71 –
0.90g of water of crystallisation lost.
Na2SO4
H2 O
Mass
0.71
0.90
Mole
(0.71)
(0.90)
142
18
0.005
0.05
(0.005)
(0.05) = 50
0.005
0.005
Ratio
1
10
Formula
Na2SO4.10H2O , p = 10
Q5(a) A volatile compound contains 40% carbon, 6.67% hydrogen and 53.3% oxygen.
(i)
Determine the empirical formulae.
(ii)
If it has a relative molecular mass of 60. Determine its molecular formula.
(C=12.0, O=16.0, H=1.0)
Answer
(i)
C
H
O
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Pivot point revision series 2011
Mass 40
6.67
Makesi mangi
53.3
Mole (40/12) (6.67/1) (53.3/16)
3.33
Ratio 1
6.67
3.33
2
1
Therefore E.F is
CH2O
(ii)
CH2O
= 12+2x1+16 = 30
(CH2O)n
= 60
Therefore 30n
= 60
= (60/30) = 2
n
Therefore M.F is
C2H4O2
(b) A sample of a compound of carbon, hydrogen and oxygen weighing 3.00g was burnt in oxygen to produce 4.40g of carbon
dioxide and 1.80g of water. Determine the values of x, y and z in the formula CxHyOZ
(C=12.0, H=1.00, O=16.0)
Answer
(b) CO2
= 12+16x2 = 44
Mass of C= (12/44 x 4.40) = 1.20g
H2O = 1x2+16 = 18
Mass of hydrogen = (2/18 x 1.80) = 0.2g
Total mass of carbon and hydrogen = (1.20+0.2)
= 1.40g
Therefore mass of oxygen from compound = (3.00 – 1.40)
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Makesi mangi
= 1.60g
C
H
O
Mass 1.20
0.2
1.60
Mole (1.20/12)
(0.2/1)
(1.60/16)
0.2
0.1
Ratio 1
2
1
Therefore x=1,
y=2,
0.1
z=1, or their multiples in the same ratio
(c) Many metal oxides are simply prepared by heating metal in air but tin oxide is prepared by first reacting tin with
concentrated nitric acid. A piece of tin foil is weighed in a test tube and then conc. Nitric acid added in drops until no more
tin is left. The liquid is then gently boiled away until evolution of brown gas is over.
Results
Mass of tube and tin oxide
= 28.17g
Mass of tube and tin = 27.63g
Mass of empty test tube
= 25.62g
(i) What is the mass of tin used?
(ii) What mass of oxygen does it combine with?
(iii) Determine mass of oxygen that combines with 119g. of tin and hence formula of oxide.
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Makesi mangi
(iv) Write specific name of tin oxide (Sn = 119, O=16.0)
Answer
(i)
Mass of tin
= 27.63
25.62
2.01 g
(ii)
Mass of Oxygen
= 28.17
27.63
0.54g of oxygen
(iii)
2.01g. of tin
= 0.54g
Therefore; 119g of tin
= (0.54/2.01 x 119)g
= 32.0g
1mole of tin combines with 2 moles of oxygen
Therefore formula of oxide is SnO2
(iv)
Tin (IV) Oxide
(d) An oxide of lead was weighed in a porcelain boat and then reduced by heating in a stream of hydrogen. The boat was
allowed to cool with hydrogen still passing over it and then it was weighed. It was re-heated in hydrogen, cooled and reweighed until a constant mass was attained for boat and lead.
Results
Mass of boat
= 10.2g
Mass of boat + lead
= 17.37g
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Pivot point revision series 2011
Final mass of boat + lead
Makesi mangi
= 16.41g
(Pb = 207, O=16)
(i) Name a drying agent for hydrogen.
(ii) Why was the (i) The boat cooled with hydrogen still passing over it?
(ii) The experiment repeated until constant mass was attained
(iii) What mass of lead was formed in this experiment
(iv) What mass of oxygen was combined with the lead in the oxide sample
(v) Calculate the mass of oxygen combined with 1 mole of lead atoms and name it.
(vi) Write equation of reaction between lead oxide and hydrogen gas.
Answer
(i)
CaCl2 or. Conc. H2SO4.
(ii)
I. To ensure oxygen is kept away from hot lead.
II. To ensure complete reaction
(iii)
Mass of lead = 16.4110.20
6.21g
(iv)
Mass of oxygen
= 17.37
16.41
0.96g
(v)
6.21g of Pb
= 0.96g of oxygen
207g of lead
= (0.96 x 207)
6.21
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Pivot point revision series 2011
Makesi mangi
= 32.0g
Mole Oxygen = 32 = 2moles
16
Combined with 1 mole of lead (207g)
Formula = PbO2, Name = Lead (IV) Oxide.
(vi)
PbO2 + 2H2(g)
Pb(s) + 2H2O(l)
(e) In another experiment 4.14 g of lead was obtained from 4.46g of another lead oxide sample.
(i) Calculate the number of moles of oxygen combined with 1 mole of lead atoms in this oxide.
(ii) Give formula and name of oxide
(Pb = 207, O = 16)
Answer
(i)
Mass of lead =
4.14g
Mass of oxygen=
4.46
4.14
0.32g
Mass of oxygen combining with 1 mole of lead
= (0.32 x 207)
4.14
= 16g
(ii)
16g
= (16/16) mole of oxygen
Formula = PbO
Name = Lead II Oxide
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Pivot point revision series 2011
Makesi mangi
(f) Calculate the mass of each of the following.
(i)
0.24mole of chlorine atoms
(ii)
0.12 mole of chlorine molecules
(iii)
0.25mole of Iron
(iv)
2 moles of magnesium
(v)
0.1 mole of nitrogen
(Cl = 35.5, Fe = 56.0, Mg = 24.0, N = 14.0)
Answer
(i)
Mass of 0.24 mole of atoms of chlorine = (0.24x35.5)
= 8.52g
(ii)
Chlorine = Cl2 molecules
= (35.5x2) = 71
Mass of 0.12 mole = (0.12x71)
= 8.52g
(iii)
Mass of Iron
= (0.25x56)g
= 14g
(iv)
Mass of 2 moles of magnesium = (24x2)
= 28g
(v)
Nitrogen = N2 molecules
= (14x2)
= 28g
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Pivot point revision series 2011
Makesi mangi
0.1mole of nitrogen = (0.1x28)g
= 2.8g
Q6. (a)The atoms of copper are arranged in the solid form in a cubic structure. The density of copper is 8g/cm3 and the
diameter of each atom is
2x10-10m. (Cu = 64.0)
(i) Find volume of 1 mole of copper
(ii) The structure is made up of unit cubic cells each containing 6 atoms and a having a width of 4.31x10-10m. How many atoms
will 1 mole of copper have.
Answer
(i)
(ii)
8g
=
1cm3
64g
=
(1/8x64)cm3
=
8cm3
Volume of unit cell
=
(4.32x10-8)3cm3
=
8.01x10-23cm3
Number of unit cells in volume of
1mole =
(8.0/8.01x10-23)
= 1.00x1023 cells
Number of atoms in 1 mole
= 6x1.00x1023
= 6.0x1023 atoms
(b) The diffusion of three gases, oxygen, gas X, and sulphur dioxide were studied. 50cm3 of Oxygen was found to diffuse
through a hole in 30 seconds. The same volume of gas X took 7.5 seconds.
(i)
Calculate the R.M.M of gas X
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Pivot point revision series 2011
(ii)
Makesi mangi
Calculate the rate of diffusion of Oxygen
(iii) How long will it take 100cm3 of sulphur dioxide to diffuse through the same hole.
(s =32.0, O = 16.0)
(iv)
What assumptions have you made?
Answer
(i)
According to Graham’s Law
rd(O2) =
Mx
under the same conditions
rd(x)
MO2
of temperature & pressure.
For equal volumes diffused rd(0z)
rd(x)
Where = rd
=
Mx = tx
Mo2 =
to2
= rate of diffusion, Mx and Mo2 are relative masses for X and O2 respectively tx and to2 are times taken for diffusions
for X and Oxygen respectively.
But. O2
= 16 x 2
= 32
Therefore
Mx = 7.5
32
Therefore
30
Mx = 7.5
32
2
 Mx =
30
7.52 x 32
302
=2
(ii)
Rate of diffusion of oxygen, rd (o2) = 50/30 = 12/3cm3/sec.
(iii)
rdO2
=
Mso2/Mo2
But SO2 = 32+ 16x2 = 64
r
d(SO2)
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Pivot point revision series 2011
12/3 / 100/tso2 =
64
 5/3 x tso2/100 = 2
/32
tso2 = 100 x  2 x 3
Makesi mangi
5
= 84.8 sec
5
(iv)
The temperature and other conditions of diffusion remained the same.
Q7(a) Define the term molarity
Answer
Moles in 1dm3 (1000cm3 of solution) of solution
(b) Calculate the molar concentration of each of the following.
(i) Solution of 4.0g of NaoH in 250cm3 solution
(ii) Solution having 0.1mole of hydrochloric acid in 500cm3 of solution
(iii) Solution prepared by having 2.00g of sodium hydroxide in 250cm3
(iv) Solution having 5.85g of sodium chloride in 500cm3
(Na = 23.0, O = 16.0, H = 1.00, Cl = 35.5)
Answer
(i)
NaOH = 23+16+1 = 40
Mole NaOH = (4.0/40) = 0.1mole in 250cm3
= (0.1/250 x 1000)
Molarity
= 0.4M
(ii)
Molarity of HCl
= (0.1/500 x 1000)
= 0.2M
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Pivot point revision series 2011
(iii)
Makesi mangi
NaOH = 23+1+16 = 40
Mole NaOH = (2/40) = 0.05 mole in 250 cm3
Molarity
= (0.05/250 x 1000)
= 0.2M
(iv)
NaCl = 23+35.5 = 58.5
Mole NaCl = (5.85/58.5) = 0.1mole NaCl in 500cm3
Molarity
= (0.1/500 x 1000)
= 0.2M
(c) What mass of;
(i)
Sodium hydroxide is in 200cm3 of its 2M solution
(ii)
Sodium in 200cm3 of 2Molar Sodium hydroxide solution
(iii) Chloride ions in 25.0cm3 of 2Molar calcium chloride solution.
(iv) Chloride ions in 2000cm3 of 2Molar Calcium chloride solution.
(Ca = 40.0, Na = 23.0, O = 16.0, H = 1.00, Cl = 35.5)
Answer
(i)
NaOH =
(23.0+16.0+1.00) = 40
2mole =
2x40 = 80g in 1000cm3
 200cm3
=
(ii)
=
16.0g
Mass of NaOH
 Mass of Na
( 80/1000 x 200)
=
16.0g (above)
=
(23/40 x 16.0)
=
9.2g
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Pivot point revision series 2011
(iii)
CaCl2
=
(40+35.5x2)
Makesi mangi
=
40.0
71.0
111.0
Mass of CaCl2 =
=
(2/1000 x 25.0) x 111
5.55g
Mass of chloride ions = (71/111 x 5.55)
(iv)
Mass of CaCl2 =
=
3.55g
(2/1000 x 2000) x 111 = 444g
Mass of chloride ions = (71/111 x 444) = 284g
(d) How many chloride ions are there in 200cm3 of 0.2moldm-3 of magnesium chloride solution ( L = 6.0x1023mol-1)
Answer
Mole of MgCl2
=
=
(0.2/1000 x 200)
0.04mole
But every MgCl2 has two Cl- ions
Mole of Clions of Cl-
=
(0.04x2)
=
0.08 mole
=
(0.08x6.0x1023)
=
4.8x1022ions
(e) 5.85g of NaCl was made into a solution of 200cm3 in water. Calculate the number of sodium ions in this solution.
(Na = 23.0, Cl = 35.5
L = 6.0 x 2023 mol-1)
Answer
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Pivot point revision series 2011
Makesi mangi
NaCl = 23+35.5 = 58.5
Mole NaCl = (5.85/58.5) = 0.1 mole in 200cm3
But in 1 NaCl there is one Na+
Hence; Mole Na+
= mole of NaCl

= (0.1x6.0x1023)
No. of Na+
= 6.0x1022 ions
Q8. (a) Define the following terms
(i)
Standard solution
(ii)
Molar mass
(iii)
A molar solution
Answer
(i)
A standard solution is one whose concentration is known accurately.
(ii)
Molar mass is the mass of 1 mole of a substance in grams
(iii)
A molar solution is one which contains 1 mole in 1000 cm3 (1dm3)
(b) Calculate the molarity of the following solutions.
(i)
2.0g of NaOH in 200cm3 of solution
(ii)
5.85g of sodium chloride in 500cm3 solution
(iii)
8.0g of sodium hydroxide in 1000cm3 solution
(Na = 23.0, O = 16.0, H = 1.00), (1 = 35.5)
Answer
(i)
NaOH = (23+16+1) = 40
Mole NaOH = (2/40) = 0.05mole
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Molarity
=
=
(ii)
NaCl =
Makesi mangi
(0.05/200 x 1000)cm3
0.25M
(23+35.5)
= 58.5
Molarity NaCl = 5.85/58.5 x 1000/500
= 0.2m
(iii)
NaOH = 23+16+1 = 40
Molarity of NaOH = (8/40)
= 0.2M
(c) Calculate the concentration in moles/dm3 (molarity) of the following solutions of sulphuric acid
(i)
0.1mole in 50.0cm3 of solution
(ii)
0.20 mole in 250cm3 of solution
(iii)
0.30 mole in 250 cm3 of solution
Answer
(i)
Molarity = (0.1/50 x 1000) = 2.0M
(ii)
Molarity = (0.2/250 x 1000) = 0.8M
(iii)
Mole/dm3 = (0.3/250 x 1000) = 1.2M
(d) Calculate the number of moles of;
(i)
NaOH given 20.0cm3 of its 1.0M solution
(ii)
Sulphuric acid given 25.0cm3 of its 2.0M solution
(iii)
Hydrochloric acid given 500cm3 of its 1.5M solution.
(iv)
Sulphuric acid given 1500cm3 of its 2.0M solution
Answer
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Makesi mangi
(i)
Moles NaOH = (1.00/1000 x 20.0)
(ii)
Moles H2SO4 = (2/1000 x 500) = 0.75moles
(iii)
Moles HCl
(iv)
Mole H2SO4 = (2/1000 x 1500) = 3.0 moles
= 0.02Mole
= (1.5/1000 x 500)
= 0.75moles
Q9. (a) Calculate the Molar gas volume if;
(i)
0.89g of chlorine occupies 281.0cm3 at standard temperature and pressure (s.t.p)
(ii)
0.525g of nitrogen occupies 420cm3 at standard temperature and pressure (s.t.p)
(iii)
From your results in (i) and (ii) what would you conclude?
(Cl = 35.5, N = 14.0)
Answer
(i)
Cl2
= (35.5x2) = 71
 Mole Cl2
= (0.89/71) = 0.01254
Therefore vol. Of 1mole = (281.0/0.01254 x 1 x 1/1000)
= 22.4dm3
(ii)
N2
= (14x2) = 28
1mole of N2 = (420 x 20 x 1) dm3
0.525
(iii)
1000
= 22.4dm3
The molar gas volume of any gas under same conditions is a constant.
(b) Sodium nitrate decomposes on heating to form sodium nitrite and oxygen. 900cm3 of gas was formed at 270C and
770mmHg.
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Makesi mangi
(i)
Write a balanced equation for reaction
(ii)
Calculate volume of gas produced at s.t.p (s.t.p = 2730K, 760mmHg)
(iii)
What mass of sodium nitrate must have been decomposed to produce this volume of gas
22.4dm3)
(Na = 23.0, N = 14.0, O = 16.0)
Answer
(i)
2NaNO3(s)
2NaNO2(s) + O2(g)
(ii)
P1V1
= P2V2
T1
T2
Where P1V1 and T1 refer to s.t.p
Conditions and P2 = 770mmHg, V2
= 900cm3 and T2 = (2730 + 270)K
V1
= P2V2 x T1
T2
P1
= (770 x 900 x 273)
300x760
= 829.8cm3
(iii)
From equation in (i) above mole ratio is 2 : 1
But mole of O2
= (829.8)
22400
 Mole of NaNO3
= 0.037 x 2
= 0.074
But NaNO3
= (23+14+16x3)
= 85
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(Molar Gas volume at s.t.p =
Pivot point revision series 2011
 Mass
Makesi mangi
NaNo3 = 0.074x85
= 6.30g
(c) (i) State Avogadro’s hypothesis.
(ii) 10.0cm3 of a gas containing only carbon and hydrogen required 30.0cm3 of oxygen for complete combustion. 20.0cm3 of
carbon dioxide were formed. All volumes were measured at the same temperature and pressure. Determine the formula of
hydrocarbon compound.
Answer
(i)
Equal volumes of gases under the same conditions of temperature and pressure contain the same number of particles
(ii)
Let formula of hydrocarbon compound be
Hence CxHy + (x + y/4) O2
10cm3 30cm3
CxHy
xCO2(g) + y/2 H2O(l)
20cm3
Volume ratio is 1:3:2
And mole ratio is 1: (x+y/4) : x
 x=2
and
x+y/4 = 3
And hence 2+y/4 = 3
 y/4 = 1
y =4
(d) What mass of carbon dioxide will be formed when 2000cm3 of carbon monoxide measured at s.t.p is completely burnt in
oxygen? (M.G.V = 22.4dm3, C = 12.0, O = 16.0)
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Makesi mangi
Answer
Equation of reaction;
2CO(g) + O2(g)
2CO2(g)
Mole ratio or volume ratio under same conditions of temperature and pressure.
= 2:2
= 1:1 (CO : CO2)
Volume of carbon dioxide formed = 2000cm3
No. of mole
= (2000/22400)
But CO2
Mass CO2
= 0.0893
= 12+16x2 = 44 (moles x relative mass)
= 0.0893x44
= 3.93g
(e) (i) State Gay-Lussac’s Law
(ii) Write equation for the burning of hydrogen in air.
(iii) If all volumes are measured at s.t.p, state volumes and names of the remaining gases when 100cm3 of hydrogen are exploded
with 250cm3 of oxygen.
Answer
(i)
Gases react in volumes that bear simple ratio’s among reactants and products.
(ii)
2H2(g) + O2(g)
2H2O(l)
Volume ratio of gases = mole ratio
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Makesi mangi
= 2:1
100cm3 of hydrogen requires only 50cm3 of oxygen
Volume of oxygen unreacted = (250-50)cm3
= 200cm3
Water will exist in solid state at 273K
(f)
Impure calcium carbonate containing 10% of an inert impurity were subjected to strong heating. If 2.5g of the solid
was heated; calculate
(i)
Mass of solid residue
(ii)
Mass and volume of gas that would be evolved at s.t.p
(M.G.V at s.t.p = 22.4dm3) (Ca = 40.0, C = 12.0, O = 16.0)
Answer
(i)
Mass of pure CaCO3 = (90% of 2.5)g = 2.25g
i) Equation
CaCO3
Mole ratio
1
CaO + CO2(g)
:
1
: 1
CaCO3 = 40+12+48 = 100
Mole CaCO3 = (2.25) = 0.0225 = mole CaO. (mole ratio is 1:1)
100
But CaO = 40+16 = 56
Mass of CaO = 0.0225 x 56
= 1.26g
Total mass of solid residue = 10 x 2.5 + 1.26
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Makesi mangi
100
= 1.51g
ii) Mole of CO2
= 0.0225
(ratio = 1:1)
But CO2 = 12+16x2 = 44
 Mass of CO2
= 0.0225 x 44
= 0.99g
Vol. Of CO2 at s.t.p
= 0.0225 x 22.4dm3 ( moles x relative mass)
= 0.504dm3 (Moles x M.G.V)
Q10. 50.0cm3 portions of a 0.5M solution of barium nitrate were added to separate samples of six solutions of 0.5M solution of a
soluble carbonate. The mass of precipitate formed in each case was determined and recorded in each case.
Volume of
carbonate
solution added
(cm3)
Mass of
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Makesi mangi
precipitate (g)
(a) Why was there no more precipitation even when more carbonate is added?
Answer
The reaction was complete.
(b) How many moles of the carbonate were just sufficient to precipitate all the barium nitrate?
Answer
Reaction was first complete when 50,0cm3 carbonate was added
= (0.5x50)
1000
= 0.025mole
(c) The concentration of the carbonate solution was found to be 5.3g/100cm3. Find the mass of 1mole of carbonate
Answer
5.3g in 100cm3
1000cm3 = (5.3x1000) = 53g
100
53g = 0.5mole
1mole = (53x2)g = 106g
(d) The metal carbonate has the formula, Y2CO3 where Y is not the true element symbol.
What is the
(i) Group
(ii) Relative atomic mass of Y?
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Makesi mangi
(C = 12.0, O = 16)
Answer
(i)
Valency of Y is (2/2) = 1 and the carbonate is soluble. Y is in group I
(ii)
Y2CO3 = 106
= 2y+12+16x3 Where y is the R.A.M of y
2y
= (106-60)
= 46
y
= 23
(e) Write a balanced equation for the precipitation of barium carbonate using the correct symbol for y.
Answer
Na2CO3(aq) + Ba(NO3)2(aq)
BaCO3(s) + 2NaNO3(aq)
Q11. Write ionic equation to illustrate
(i)
Precipitation of a carbonate of lead
(ii)
Reaction between a solution of carbonate salt and a dilute acid.
(iii)
Reaction between a solid carbonate and a dilute acid
(iv)
Precipitation of an iodide of lead
(v)
Precipitation of a chloride of silver
(vi)
Precipitation of Iron III hydroxide.
(vii)
Reaction between sodium sulphite and dilute acid
(viii)
Precipitation of a sulphate of barium
(ix)
Reaction between a hydrogen carbonate and a dilute acid
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Makesi mangi
(x)
Reaction between a dilute acid and the metal, zinc
(xi)
Reaction between lead and a solution of silver salt.
(xii)
Neutralisation
(xiii) Reaction between dilute alkaline solution and bromine
(xiv)
Reaction between dilute sodium hydroxide solution and amphoteric hydroxide
(xv)
Chlorine and a bromide salt solution
Answer
(i)
Pb2+(aq) + CO32-(aq)
PbCO3
(ii)
2H+(aq) + CO32-(aq)
H2O(l) + CO2(g)
(iii)
CO32-(s) + 2H+(aq)
H2O(l) + CO2(g)
(iv)
Pb2+(aq) + 2I-(aq)
PbI2(s)
(v)
Ag+(aq) + Cl-(aq)
AgCl(s)
(vi)
Ba2+(aq) + SO42-(aq)
BaSO4(s)
(vii)
Fe3+(aq) + 3OH-(aq)
Fe(OH)3(s)
(viii)
SO32-(aq) + 2H+(aq)
SO2(g) + H2O(l)
(ix)
HCO3-(aq) + H+(aq)
H2O(l) + CO2(g)
(x)
Zn(s) + 2H+(aq)
H2(g) + Zn2+(aq)
(xi)
2Ag+(aq) + Pb(s)
Ag(s) + Pb2+(aq)
(xii)
H+(aq)+ + OH-(aq)
(xiii) Br2(g) + 2OH-(aq)
H2O(l)
OBr-(aq) + H2O(l) + Br-(aq)
(xiv)
2OH-(aq) + Pb(OH)2(s)
Pb(OH)4-(aq)
(xv)
Cl2(g) + 2Br-(aq)
2Cl-(aq) + Br2(g)
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Makesi mangi
(b) In each of the reactions given in (a) in (i-xv), give appropriate examples to enable you to write the full chemical equation.
Answer:
(i)
Na2CO3(aq) + Pb(NO3)2(aq)
PbCO3(s) + 2NaNO3(aq)
(ii)
(NH4)2CO3(aq) + 2HCl(aq)
2NH4Cl(aq) + CO2(g) + H2O(l)
(iii)
PbCO3(s) + 2HNO3(aq)
Pb(NO3)2(aq) + CO2(g) + H2O(l)
(iv)
Pb(NO3)2(aq) + 2KI(aq)
PbI2(s) + 2KNO3(aq)
(v)
AgNO3(aq) + NaCl(aq)
NaNO3(aq) + AgCl(s)
(vi)
BaCl2(aq) + K2SO4(aq)
BaSO4(s) + 2KCl(aq)
(vii)
FeCl3(aq) + 3NaOH(aq)
Fe(OH)3(s) + 3NaCl(aq)
(viii)
Na2SO3(aq) + 2HCl(aq)
2NaCl(aq) + H2O(l) + SO2(g)
(ix)
NaHCO3(s) + HCl(aq)
NaCl(aq) + CO2(g) + H2O(l)
(x)
Zn(s) + H2SO4(aq)
ZnSO4(aq) + H2O(l)
(xi)
Pb(s) + 2AgNO3(aq)
2Ag(s) + Pb(NO3)2(aq)
(xii)
2NaOH(aq) + H2SO4(aq)
Na2SO4(aq) + H2O(l)
(xiii) Br2(g) + 2NaOH(aq)
NaOBr(aq) + NaBr(aq) + H2O(l)
(xiv)
NaOH(aq) + Al(OH)3(s)
NaAl(OH)4(aq)
(xv)
Cl2(g) + 2NaBr(aq)
Br2(g) + 2NaCl(aq)
(c) A slight excess of copper sulphate solution was reacted with 0.327g of pure Zinc powder in order to be able to determine
the ratio of reaction between copper ions and Zinc. A red brown solid was formed. The solid was filtered, washed with
water and finally acetone and then warmed to dry and weigh.
(i)
What was the red brown solid?
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Makesi mangi
(ii)
How many moles of Zinc are in 0.327g of metal (Zn = 65.4) ?
(iii)
Why was an excess of copper sulphate solution used?
(iv)
The mass of the red brown solid obtained was 0.3175g. Calculate the moles of copper formed (Cu = 63.5)
(v)
The charge on the ion of Zinc? Explain your answer.
(vi)
Write ionic equation of reaction between Zinc metal and copper sulphate solution.
Answer
(i)
Copper
(ii)
Mole of Zinc = (0.327) = 0.005 mole
65.4
(iii)
To ensure complete reaction of Zinc
(iv)
Mole of copper = (0.3175/63.5) = 0.005 mole
(v)
Charge on ion of Zinc = +2
No. of mole of copper ions displaced = no. of mole of Zinc ions formed. The number of electrons transferred is the same and
this implies each ion must have the same charge since the ratio of moles is 1:1
(vi)
Zn(s) + Cu2+(aq)
Cu(s) + Zn2+(aq).
Q12. The diagram below is a set up of an experiment to study the molar gas volume of carbon dioxide.
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Makesi mangi
Piston
Carbon dioxide
Boiling tube
Ignition tube
Sulphuric acid
Sodium carbonate
A sample of 0.53g of anhydrous sodium carbonate was weighed in ignition tube and then put in acid, in excess, in a boiling tube
which was then corked tightly and connected to a graduated syringe. Volume of carbon dioxide formed was noted at the room
temperature conditions as the reaction occurred when boiling tube was tilted and gas evolved.
(a) Write equation of reaction that occurred.
Answer
Na2CO3(s) + H2SO4(aq)
Na2SO4(aq) + CO2(g) + H2O(l)
(b) Why was excess acid used?
Answer
Excess acid was used to ensure complete reaction of the sodium carbonate.
(c) Calculate the moles of sodium carbonate reacted and hence the moles of carbon dioxide.(Na = 23.0, C = 12.0, O = 16.0)
Answer
Na2CO3 = 23x2+16x3 = 106 mole Na2CO3 = 0.53/106
= 0.005 mole
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Pivot point revision series 2011
But from (a) mole ratio is 1:1
Makesi mangi
mole CO2 = 0.005.
(d) Calculate the molar gas volume of carbon dioxide at the room conditions if 119cm3 of gas was collected.
Answer
0.05 mole
CO2(g) = 119cm3
CO2(g) = (119/0.005) cm3
1mole
= 23800cm3
= 23.8cm3
(e) The molar gas volume for hydrogen under these conditions has been found to be more than 24.0dm3. Suggest a reason for
this.
Answer
At low temperatures the intermolecular attraction of high relative molecular mass are significant and pull the molecules reducing their
pressure at collision with the walls. For a smaller molecule like hydrogen the attractive forces (Van der Waal) forces are less hence H2
registers a larger volume as particles are influenced less by attractive forces.
Q13. (a) 1.0g of pure calcium carbonate was added to 60.0cm3 of 0.1M HCl. What mass of carbonate remained unchanged?
(H = 1.0, Cl = 35.5), Ca = 40.0, C = 12.0, O = 16.0)
Answer
Equation:
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)
Mole ratio is 1:2
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Makesi mangi
But mole HCl = (0.1/1000x60.0)
= 0.006 mole
Mole CaCO3 required = (0.006)
= 0.003
2
CaCO3 = (40+12+16x3) = 100
Mass CaCO3 reacted = (0.003x100)
= 0.3grams
Mass of CaCO3 unreacted
= (1.00-0.30)g
= 0.7grams
(b) What volume of 0.8M sulphuric acid can be made from 1dm3 of 1.0M H2SO4?
Answer
1.0M acid
and 0.8M acid
(1.0-0.8)
1.0mole in 1000cm3
0.8mole in 1000cm3
= 0.2mole
to make 0.8M dissolve 0.2mole in (1000/0.8 x 0.2)
= 250cm3
Total volume = 1250cm3 i.e. add
250.0cm3 of water to the 1dm3 of 1M solution
OR To make 0.8M from 1000cm3 of 1M solution, volume
= (1000/0.8x1) = 1250cm3
Q14. (a) Calculate the molarity of hydrochloric acid, 25.0cm3 of which exactly neutralises or reacts with.
(i)
20.0cm3 of 0.1M Na2CO3 solution
(ii)
30.0cm3 of 0.2M KOH
(iii)
20.0cm3 of 0.1M aqueous ammonia solution.
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Makesi mangi
Answer
(i)
Mole Na2CO3 = 0.1 x 20
1000
= 0.002 mole
Equation: Na2CO3(aq) + 2HCl(aq)
Ratio is
Mole HCl
2NaCl(aq) + CO2(g) + H2O(l)
1:2
= (0.002x2) = 0.004
0.004 mole HCl in 25.0cm3
Molarity of HCl
=
(0.004 x 1000)
25
= 0.16M
(ii)
Mole KOH = (0.2 x 30)
1000
= 0.006Mole
Equation:
KOH(aq) + HCl(aq)
KCl(aq) + H2O(l)
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Makesi mangi
Mole ratio is 1:1
= 0.006 in 25.0cm3
Mole HCl
Molarity
= (0.006 x 1000)
25
(iii) Mole NH3(aq)
= 0.24M
= (0.1 x 20)
1000
= 0.002mole
Equation: NH4OH(aq) + HCl(aq)

Hence:
NH4Cl(aq) + H2O(l)
ratio is 1:1
mole
HCl = 0.002 in 25cm3
Therefore : Molarity = (0.002 x 1000)
= 0.08m
25
(b) 30.0cm3 of 1Molar potassium hydroxide solution was exactly neutralised by 7.5cm3 of sulphuric acid solution. Calculate the
molarity of sulphuric acid.
Answer
Moles KOH =
(1/1000 x 30)
= 0.03mole
Equation:
2KOH(aq) + H2SO4(aq)
Mole ratio
is
K2SO4(aq) + H2O(l)
2:1
Mole H2SO4 = (0.03/2) = 0.015 mole in 7.5cm3
Molarity of H2SO4
= (0.015 x 1000)
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Makesi mangi
7.5
= 2.0M
(c) Calculate the molarity of potassium hydroxide solution if its 20.0cm3 solution reacts exactly with;
(i)
20.0cm3 of 1.2M H2 SO3
(ii)
15.0cm3 of 1M HNO3
(iii)
25.0cm3 of 0.2M phosphorus (V) acid, H3PO4
Answer
(i)
Mole H2SO4 = (1.2/1000 x 20 ) = 0.024
Equation:
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
Mole ratio is 1: 2
Mole KOH = (0.024x2) = 0.048mole in 20.0cm3
Molarity of KOH
= (0.048 x 1000)
20
(ii)
Mole HNO3 = ( 1 x 15)
1000
Equation:
= 0.015mole
HNO3(aq) + KOH(aq)
KNO3(aq) + H2O(l)
Mole ratio is
1:1
Mole of KOH
= 0.015 in 20.0cm3
 Molarity of KOH
= (0.015 x 1000)
20
= 0.75M
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(iii)
Makesi mangi
Mole of phosphorus (V) acid = (0.2 x 25)
1000
= 0.005 mole
Equation:
H3PO4(aq) + 3KOH(aq) K3PO4(aq) + 3H2O(l)
Mole ratio
1
Mole KOH
= (0.005x3)
:
3
= 0.015 in 20.0cm3
Molarity
= (0.015 x 1000)
20
= 0.75m
(d) Calculate the volume of 0.1M sulphuric acid that will react exactly with.
(i) 25.0cm3 of 0.1M NaOH
(ii) 20.0cm3 of 0.2M Na2CO3
(iii) 30.0cm3 of 0.3M NH3(aq)
Answer
(i) Mole NaOH = (0.1 x 25) = 0.0025mole
1000
Equation:
2NaOH(aq) + H2SO4(aq)
Na2SO4(aq) + 2H2O(l)
Mole ratio is 2:1
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Mole H2SO4 = 0.0025
Makesi mangi
= 0.00125mole
2
But molarity of H2SO4
= 0.1
Volume of H2SO4(aq) = (1000 x 0.00125)
0.1
= 12.5cm3
(ii) Mole of Na2CO3 = (0.2x20)
1000
= 0.004 mole
Equation
Mole ratio
Na2 CO3(aq) + H2SO4(aq)
1
mole H2SO4
:
Na2SO4(aq) + H2O(l) + CO2(g)
1
= 0.004
But molarity of H2SO4 = 0.1
 Volume of Sulphuric acid = (1000 x 0.004)
0.1
= 40.0cm3
(iii) Moles of NH3(aq) = (0.3 x 30)
1000 = 0.009mole
Equation:
2NH4OH(aq) + H(s)SO4(aq)
Or
2NH3(aq) + H2SO4(aq)
(NH4)2SO4(aq) + 2H2O(l)
(NH4)2SO4(aq)
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Makesi mangi
Mole ratio is 2 : 1
Mole of H2SO4
= 0.0045
But molarity of H2SO4
Vol. Of H2SO4(aq)
= 0.1
= (1000/0.1 x 0.0045)
= 45.0cm3
Q15 (a) A certain carbonate reacts with an acid:
XCO3(s) + 2HNO3(aq)
X(NO3)2(aq) + H2O(l) + CO2(g)
When 2.95g of the carbonate is dissolved in 50.0cm3 of 2MHNO3, the remaining acid requires 30.0cm3 of 1M NaOH to neutralise.
(i)
Calculate the moles of nitric acid in 50.0cm3
(ii)
Calculate the moles of sodium hydroxide in 30.0cm3
(iii)
Calculate the relative atomic mass of X and hence identify element.
(C = 12.0, O = 16.0)
Answer
(i)
Moles of nitric acid
= (2.0/1000 x 50) = 0.1mole
(ii)
Moles of NaOH
= (1.0/1000 x 30) = 0.03mole
= moles HNO3 (excess) as ratio is 1 : 1
NaOH(aq) + HNO3(aq) NaNO3(aq) + H2O(l)
(iii)
From equation mole ratio is 1:2
But moles of HNO3 reacted = (0.1 – 0.03)
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Makesi mangi
= 0.07moles
Moles of XCO3 = (0.07/2) = 0.035moles
0.035moles XCO3
= 2.0g
1mole XCO3 = (2.95/0.035) = 84.3
But CO3 = 12+16x3 = 60
R.A.M. of X = (84.3 – 60)
= 24.3
Element X is magnesium.
(b) 4.00g of magnesium carbonate having some magnesium sulphate impurity was dissolved in 250cm3 of 1M HCl. 25.0cm3 of
the resulting solution was found to require 28.8cm3 of 0.6M NaOH to neutralise.
(i)
Calculate the number of moles of sodium hydroxide in 28.8cm3
(ii)
Calculate the number of moles of hydrochloric acid in 250cm3 of solution of the 1M HCl.
(iii)
Determine the mole of HCl in 250cm3 solution mixture
(iv)
Calculate the number of moles of MgCO3 in mixture.
(v)
Calculate the mass of magnesium sulphate impurity and hence its percentage composition. (Mg = 24.3, C = 12.0, O = 16.0)
Answer
(i)
Moles of NaOH = (0.6/1000 x 28.8) = 0.01728 moles
(ii)
Moles of HCl = (1/1000 x 250) = 0.25moles
(iii)
Equation:
NaOH(aq) + HCl(aq)
Mole ratio
=1
:
NaCl(aq) + H2O(l)
1
Moles of HCl = moles of NaOH
25.0cm3 of mixture = 0.01728 ( see (i) above)
 250cm3of mixture = (0.01728 x 250) = mole of HCl = 0.1728
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25
(iv)
Mole HCl in original solution = 0.25
Mole HCl reacted with MgCO3 = (0.25-0.1728)
Equation: MgCO3(s) + 2HCl(aq)
Mole ratio
= 0.0772
MgCl2(aq) + CO2(g) + H2O(l)
=2 : 1
Mole of MgCO3
= (0.0772)
= 0.0386mole
2
(v)
MgCO3
= 24.3+12+16x3 = 84.3
Mass of MgCO3
= (0.0386x84.3) = 3.25g
Mass of MgSO4
= (4-3.25)
= 0.75g
Percentage composition (impurity) = 0.75 x 100% = 18.8%
4.00
(c) 40.0cm3 of a solution of an alkali, MOH was found to have 0.12g of solute. It was exactly neutralised by 25.0cm3 of 0.2M HCl.
(i) Write ionic equation for the reaction
(ii) Calculate the relative formula mass of MOH and hence the relative atomic mass of M (O = 16.0, H = 1.0)
(iii) Calculate the concentration of alkaline solution in mole dm-3
Answer
(i)
H+(aq) + OH-(aq)
(ii)
Mole of HCl reacted = (0.2/1000 x 25)
H2O(l)
= 0.005mole
But from (i) above ratio of moles of HCl : MOH = 1:1
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Mole of MOH = 0.005
0.12g = 0.005mole
1.0moles
= (0.12/0.005 x 1)
= 24
R.F.M of MOH = 24
and MOH
m
= m+16+1
= (24-17)
r.a.m of M
Where m = r.a.m of M
=7
=7
Molar concentration = (0.005/40 x 1000)
(iii)
= 0.125M
PRACTICE QUESTIONS
Q1 (a) Explain the following observations in terms of particulate nature of matter.
(i) When a bottle of perfume is opened at one corner of a room, the smell soon spreads throughout the room.
(ii)When white crystals of ammonium chloride are heated steamy fumes are produced which reform a white solid.
(b) In an experiment a long glass tube of length of 180cm and a diameter of 4cm was clamped horizontally. A pad of cotton wool was
soaked in concentrated hydrogen sulphide (gives off hydrogen sulphide) and a pad was soaked in concentrated sulphur dioxide
solution (gives off SO2 gas) were placed simultaneously at opposite ends of the glass tube. The temperature of the apparatus was
maintained at 250C. After 27 minutes a faint yellow ring was formed inside the tube as shown.
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180-x
x
Cotton wool
cotton wool
Soaked in Conc. hydrogen sulphide
soaked in Conc. sulphur
dioxide solution
(i) Explain what happens in the tube.
(ii) Name solid forming the yellow ring
(iii) According to Graham’s Law
MH2s = V2so2
Mso2
V2H2s
Where MH2s and Mso2
are relative molecular masses
For hydrogen sulphide and sulphur
Dioxide and VH2s
And Vso2 are their velocities respectively.
Determine the value of x. (S = 32.0, O = 16.0, H = 1.00)
(iv) State another expression of Graham’s Law of diffusion.
(v) How would each of the following affect the formation of ring.
I. Performing the experiment at 350C instead of 250C.
II. Holding the tube in a vertical position with the hydrogen sulphide
at the lower end.
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(vi) Write equation for reaction producing the yellow ring as sulphur dioxide oxidises hydrogen sulphide to form water and the
yellow solid.
(c) In a similar experiment as in (b) above cotton wool was soaked in concentrated ammonia and concentrated hydrogen bromide at
opposite ends. This time a white band was observed in the same tube.
(i) Name the white solid
(ii) Write an equation for the reaction that produces the white solid.
(iii) Calculate
I.
The distance from the band to each end of the tube.
II. The time it takes for the band to form. Use other appropriate data from (b) above
(H = 1.00, N = 14.0, Br = 80.0)
Q2. (a) Explain each of the following in terms of Kinetic theory:
(i) Diffusion of a gas
(ii) Evaporation of a liquid
(iii) Dissolving a solute in solvent
(b) (i) State Boyle’s Law
(ii) State Charles’ Law
(iii) Explain the gas laws, Boyle’s Law and Charles’ Law in terms of Kinetic Theory.
(c) The volume of a given mass of gas is 300cm3 at 400C and 650mmHg
(i) What will be the volume at s.t.p.
(ii) What will be the volume of the same mass of gas at the same temperature when the pressure is increased to 760mmHg?
(iii) What will be the pressure when the volume is 250cm3 at the same temperature?
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(d) A certain mass of gas occupies 200cm3 at s.t.p. What volume would it occupy at;
(i) 800C and 750mmHg
(ii) –710C and 730mmHg
Q3. (a) Thallium is isotopic with a composition of ;
203
TI
accounting for its 29.52% in nature while
TI
is 70.48%
81
205
81
(i) Calculate the Relative atomic mass of Thallium.
(ii) Given 10,000 atoms of thallium how many atoms would there be of each isotope?
(b)
The relative atomic mass of chlorine is 35.5. Determine the percentage isotopic composition of chlorine. It has isotopes
35
Cl
17
and
37
(Clue. Let 35Cl be x% abundant)
Cl
17
17
(c) For the isotope of chlorine 35Cl
17
(i) Give atomic number
(ii) Give mass number
(iii) Give the number of neutrons
(iv)
Draw the atomic structure
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(v) Write the electronic configuration.
(d) Determine the relative formula mass or molecular mass of
(i) Fe(OH)3
(ii) SO3
(iv)
(iv) SOCl2
Fes(SO4)3
(iii) Cl2O
(vi) FeCl3
(Fe = 56.0, O = 16.0, H = 1.00, Cl = 35.5, S= 32.0)
(e)
Calculate the;
(i)
Number of moles of;
I.
Phosphorus in 15.5g of its solid
II. Chlorine atoms in its 17.75 grams
III. Chlorine molecules in 35.5g of gas
IV. Chlorine molecules in 1.5x1023 atoms.
V. Chlorine atoms in 1.5x1023 atoms
(P=31, Cl = 35.5, L = 6.0x1023mole-1)
(ii)
Number of particles of;
I.
Magnesium in its 2.43g
II.
Sulphur in 0.4 moles
III.
Oxygen molecules in 16 grams
IV.
Oxygen atoms in 16 grams
V.
Magnesium in 0.02 moles of the metal.
(Mg = 24.3, O = 16, L = 6.0x1023mole-1)
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(iii)
Makesi mangi
Calculate the mass of;
I.
6x1023 molecules of chlorine, Cl2
II.
6x1023 molecules of phosphorus, P4
III.
One molecule of chlorine
IV.
0.002 moles of chlorine molecules
V.
0.04 moles of aluminium
(L = 6.0x1023, Cl = 35.5, P = 31.0, Al = 27)
3(a)
Calculate the percentage of the stated substance in the hydrated compound, CuSO4.5H2O.
(i)
Copper
(ii)
Sulphur
(iii)
Oxygen
(iv)
Water
(Cu = 63.5, S=32.0, H=1.00, O=16.0)
(b) Given the following data determine the empirical formula in each case of given composition by mass.
(i)
9.5g of magnesium and 63.92g of bromine
(ii)
Sodium 32.4%, sulphur 22.5% and oxygen 45.1%
(iii)
5.4g of aluminium and 4.8g of oxygen
(Mg = 24.3, Br = 79.9, Na = 23.0, S=32.0, O = 16.0, Al = 27.0)
(c)
Blue hydrated crystals of mass 2.495g was heated gently to form a white powder of mass 1.595g. Determine the value of x in
the formula;
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MSO4. XH2O. (M = 63.5, S=32.0, O=16.0, H=1.0)
(d)
1.35g of aluminium was heated in a stream of chlorine to produce 4.01g of aluminium chloride.
(i) Determine the empirical formula
(ii) The vapour from this compound has a relative molecular mass of 133.5.
Determine the molecular formula.
(iii) What type of bond is found in aluminium chloride? Explain.
Q4. (a) A clean crucible and lid were weighed and clean magnesium ribbon was put in it and burnt on a pipe-clay triangle by first
heating gently then strongly. The lid is lifted with tongs and then replaced before smoke escapes. The procedure was repeated
until no more change.
Lid
Crucible
Magnesium ribbon
Tripod stand
heat
Results
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Mass of crucible + lid
= 14.52g
Mass of crucible, lid and magnesium
= 15.72g
Mass of product after burning, crucible and lid
= 16.52
(i)
Determine the empirical formula of magnesium oxide.
(ii)
Why is the lid lifted and then quickly replaced?
(b) In another experiment it was found 2.0g of calcium combines with 3.55g of chlorine.
(i) Determine empirical formula
(ii) Suggest a suitable arrangement for the experiment
(iii) What is the appearance of product calcium chloride?
(iv) Write equation of reaction between calcium and chlorine.
(c) Below is a table of data on various compounds. Use the information to determine formula. The information relates to mass.
(i) C = 80%
H = 20%
R.m.m = 30
(ii) Hydrogen = 5.9%, Oxygen = 94.1% and R.m.m = 34.
(iii) Carbon is 38.75%, hydrogen = 16.1%, nitrogen = 45.2%
(iv) Hydrocarbon with 92.3% carbon and a R.m.m of 78.
(d) 2 volumes of hydrogen fluoride gas, HF(g) , combines with 1 volume of a gas X to form 2 volumes of a gas Y as the only product.
Gas X has an empirical formula NF and its r.m.m is 66.
(i) What is the molecular formula of gas X?
(N = 14.0, F = 19.0)
(ii) Give a suitable name for X
(iii) Write an equation for the reaction between HF and gas X using the correct formula for X.
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5(a) Sodium atoms in its crystal form arrange themselves in a simple cubic structure known as body centred cubic structure. Below is a
diagram of the simplest repeating unit in the lattice known as the unit cell-structure.
Sodium atom
(i) Each edge of the unit cell has a length of 4.29x10-10m. Calculate the volume of the unit cell.
(ii) The unit cell has effectively two atoms of sodium. Show how this number is arrived at.
(iii) The unit cell of two atoms of sodium has a mass of 7.64x10-23g.
I. Calculate the number of atoms of sodium in 1 mole (Na = 23)
II. Considering the length of each edge given above, Calculate the volume of one mole of sodium.
(iv) Given also that the density of sodium is 0.97gcm-3. Calculate the volume of 1mole of sodium (Na = 23)
(b) The density of hydrogen gas at 250C and 1 atmosphere is 0.08gdm-3. That of carbon monoxide under the same conditions is
1.145gdm-3. Calculate the molar gas volume of each gas. Explain the difference.
(H = 1, C = 12.0, O = 16.0)
(c) If 0.1811g of carbon dioxide occupies 100cm3. Calculate the molar gas volume of carbon dioxide under the same conditions.
(C = 12.0, O = 16)
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(d) (i) State Avogadro’s Law
(ii)In order to determine the relative proportion of ethyne and ethene in a mixture of the two, 10cm3 of mixture was exploded with
30cm3 of oxygen (an excess). After absorbing the residual CO2 in potassium hydroxide the uncombined oxygen occupied 2cm3
I.
Write equation for combustion of each gas.
II. Write equation of reaction between potassium hydroxide and carbon dioxide.
III. Determine the composition of gas mixture by volume (Clue. Let volume of one gas be xcm3 and find total volume of
oxygen in terms of x)
Q6(a) (i)
(ii)
Write equation for action of heat on sodium hydrogen carbonate.
Calculate the mass of residue when 63g of NaHCO3 was heated until a constant mass obtained.
(iii) Calculate the volume of carbon dioxide evolved at s.t.p.
(M.G.V at s.t.p = 22.4dm3)
(b) When 33.1g of lead nitrate is decomposed by heating lead oxide. Nitrogen dioxide and oxygen are formed.
(i) Write equation for the decomposition
(ii) Calculate the mass of residue
(iii) Calculate the volume of oxygen liberated at room temperature and 1 atmosphere
(M.G.V. at R.t.p. = 24.0dm3)
(c) Write equations for the following reactions;
I. Action of water on lithium nitride to produce ammonia
II.
Action of Zinc on sulphuric acid to produce hydrogen gas
III. Action of heat on potassium chlorate V to produce oxygen and potassium chloride.
(ii) When 3.27g of Zinc was reacted with an excess of sulphuric acid at s.t.p. conditions. What volume of hydrogen gas was produced?
(M.G.V. at s.t.p = 22.4dm3).
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Q7(a) Complete the following tables in your book correctly of relative
(i)
Element
Symbol
Molar Mass
Sodium
Chlorine
Iron
Aluminium
Fluorine
19.0g
Bromine
Br
79.9g
(Na = 23.0, Cl = 35.5, Fe = 56, Al = 27.0, F = 19.0)
(ii)
Element
Mass in grams No. of moles
Magnesium
0.729
Hydrogen
0.25
Carbon
30
Zinc
327
(Mg = 24.3, H = 1, C = 12, Zn = 65.4)
No. of atoms
(L = 6.0x1023mole-1)
(iii)
Element
Mass in grams
Zinc
1.308
No. of atoms
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Bromine
15.98
Nickel
298.5
Sodium
0.46
(Zn = 65.4
Br = 79.9
Makesi mangi
N = 58.7
Na = 23)
(L = 6.0x1023mole-1)
(b) If 1 mole of nitrogen is 6.02x1023 molecules;
(i) How many atoms are in 2 moles of oxygen molecules?
(ii)
What is the mass in grams of an atom of
I.
silver
II.
hydrogen
III.
aluminium
(Ag = 108,
(c)
H = 1.00,
Al = 27.0)
A metal oxide, MO was reduced in a stream of hydrogen. 1.0g of water was formed from 4.44g of oxide.
(i) Name any two oxides that can be reduced by hydrogen
(ii)Write equation for the reduction of MO.
(iii) Calculate the mass of M that combines with 1 mole of oxygen
(M = 63.5, H = 1.0, O = 16)
(d) Calculate the mass of element which combines with 16 grams of oxygen if;
(i)
88.8%
(ii) 47.6% by mass of oxide is oxygen
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(e) 2.0g of a mixture of iron and oxide of iron reacted with dilute hydrochloric acid. The gas produced measured at 270C and
740mmHg was 200cm3
(i) Write equations for reactions that occurred
(ii) Find volume of hydrogen gas produced at s.t.p;
(s.t.p = 760mmHg and 2730K)
(iii) Calculate the percentage of iron in the mixture (M.G.V. at s.t.p is 22.4dm3).
Q8. (a) In each of the solutions below calculate mass of solute in:
(i) 20.0cm3 of 1M NaOH
(ii) 25.0cm3 of 2M H2SO4
(iii) 2500cm3 of 0.2M H2SO4
(iv) 5dm3 of 0.1M HCl
(v) 2dm3 of 2M H2SO4
(Na = 23.0, O = 16.0, H = 1.0, S = 32.0, Cl = 35.5)
(b) Potassium iodide was reacted with lead nitrate to study the ratio of their reaction in small test tubes of uniform cross sectional area.
2M solution of each reagent was used.
Vol. Of
4.0cm3
KI(aq)
Vol. Of
Pb(NO3)2(a
0.5cm3
4.0cm
4.0cm
3
3
1.0cm
1.5cm
3
3
4.0cm3
2.0cm3
4.0cm
4.0cm
3
3
2.5cm
3.0cm
3
3
q)
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Height of
precipitati
0.15cm
0.30c
0.45c
m
m
Makesi mangi
0.6cm
0.6cm
0.6cm
on in test
tube
(i)
Why does the height of precipitate reach a maximum even when more lead nitrate is added?
(ii)
What volume of 2M Pb(NO3)2 solution exactly reacts with 4.0cm3 of potassium iodide.
(iii) Calculate the moles of;
I.
KI used in each test tube
II. Pb(NO3)2 in 2.0cm3 solution
III. PbI2 precipitated by the 4.0cm3 of 1M KI.
(iv)
Write ionic equation of reaction for the precipitation.
(c) In another experiment 2M AgX salt solution was used. X is not a true symbol. This time 4.0cm3 of the AgX solution was the
minimum volume producing maximum height of iodine precipitate
(i) What is the charge on silver ion? Write formula of the ion
(ii) Write formula of precipitate
(iii) Write ionic equation for precipitation
(iv) Suggest a possible anion, X and write its formula.
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Q9. The hydrogen carbonate of a divalent metal M reacts with hydrochloric acid in accordance with the following equation:
MCO3(s) M(s) + 2HCl
MCl2(aq) + H2O(l) + CO2(g)
In an experiment to find the relative atomic mass of M, 2.5g of the carbonate reacted exactly with 100cm3 of 0.5M hydrochloric
acid.
(a) How many moles of hydrochloric acid are in 100cm3 of the acid solution.
(b) How many moles of the carbonate was reacted?
(c) Calculate the relative formula mass of MCO3 and hence m, the r.a.m. of M.
Q10. (a)
(C = 12.0, O = 16.0)
25.0cm3 of a solution of hydrochloric acid of 0.1M concentration reacts exactly with 21.5cm3 of a solution of sodium
hydroxide. Calculate the concentration of the sodium hydroxide solution.
(b) What volume of 0.2M sodium hydroxide reacts with;
(i) 20.0cm3 of 0.05M sulphuric acid
(ii) 25.0cm3 of 0.08M hydrochloric acid
(iii) 30cm3 of 0.08M nitric acid.
(c) Determine the concentration (in mol/dm-3) of the following.
(i) Sulphuric acid, 20.0cm3 of which neutralises 30.0cm3 of potassium hydroxide solution of 0.2M concentration.
(ii) Nitric acid 12.0cm3 of which reacts exactly with 25.0cm3 of a solution of carbonate of concentration 0.25M.
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(d) The following is an experiment that was done to determine R.a.m of X in the formula H2X.2H2O
Solutions
BA10 - a solution prepared by dissolving 6.3g of hydrated crystals H2X.2H2O in 1 litre.
BA11 – a solution prepared by dissolving 8.0g of sodium hydroxide in 1 litre.
25.0cm3 of acid solution was pipetted and then titrated with the sodium hydroxide solution. The table below shows results.
EXP
ROUGH
1ST
2ND
3RD
Final burette
13.0
25.5
38.0
33.0
Initial burette 0.0
13.0
25.5
20.0
12.5
12.5
13.0
reading (cm3)
reading (cm3)
Difference
13.0
(cm3)
(i) Calculate the molarity of sodium hydroxide solution, BA11
(ii) Write equation of reaction between the dibasic acid H2X and sodium hydroxide solutions.
(iii) Determine the molar concentration of BA10 solution.
(iv) Calculate the R.a.m. of X.
(H = 1.0, O = 16.0)
(a) What mass of sulphuric acid, H2SO4 will be required to exactly neutralise 25.0cm3 of 0.2M NaOH solution? (H = 1.0, S = 32.0, O
= 16.0)
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(b) Describe how you would prepare hydrated crystals of copper II sulphate (CuSO4.nH2O) using copper (II) carbonate and dilute
sulphuric acid as the starting materials. What mass of carbonate would be used to obtain 0.01 mole of CuSO4.nH2O (Cu = 63.5, C
= 12.0, O = 16.0)
(ii) The hydrated crystals were heated gently and the loss in mass was 36%
I. Describe observations when hydrated crystals are heated gently until no further change.
II. Determine the value of n in the formula, CuSO4.nH2O.
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CHAPTER 3
NITROGEN AND COMPOUNDS
Solved problems
1. (a) Nitrogen reacts with magnesium and sodium at high temperatures.
(i) Write equations of reactions.
(ii)The products effervesce with water. Write equations of reactions.
Answer
(i)
(ii)
3Mg(s) + N2(g)
Mg3 N2(s)
6Na(s) + N2(g)
2Na3 N(s)
Mg3 N2(s) + 6H2O(l)
3Mg(OH)2(s) + 2NH3(g)
Na3N(s) + 3H2O(l)
3NaOH + NH3(g)
(b) Oxygen is the active part of air while nitrogen is inactive
(i) Explain this statement
(ii) Why is nitrogen inactive?
Answer
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(i)
Makesi mangi
Oxygen supports combustion and most substances burn in oxygen. Nitrogen does not burn and most substances do not burn in
the gas.
(ii)
The triple bond in nitrogen, N  N is very strong and requires a lot of energy to dissociate.
(c) The diagram shows an arrangement for preparation and collection of nitrogen.
Combustion
tube
Copper turnings
N2(g)
heat
water
Sodium hydroxide solution
Trough
(i) What is the purpose of
(i) Sodium hydroxide
(ii)Copper turnings
(iii) Write equation for reactions that occur in the sodium hydroxide solution
(iv)
State any two impurities in the gas collected.
(v) What does the method of gas collection tell you about nitrogen.
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(vi)
Makesi mangi
Suggest how the gas could first be dried and then be collected.
Answer
(i) I. Sodium hydroxide is used to absorb CO2(g) from the air
II. Copper turnings are used to remove oxygen from the air
(ii)2NaOH(aq) + CO2(g)
2Cu(s) + O2(g)
Na2CO3(aq) + H2O(l)
2CuO(s)
(iii) Two impurities will be noble gases (e.g. argon) and water vapour
(iv)
The gas can be dried by first passing through conc. Sulphuric acid and then collecting by upward delivery over air.
(d) Two chemicals are usually stocked in the laboratory and may be used to produce nitrogen by heating.
(i) Name the two chemicals
(ii) Write equations of reactions that occur to produce nitrogen
(iii) Why is ammonium nitrite not stored in the laboratory.
Answer
(i) Sodium nitrite and ammonium chloride.
(ii) NaNO2(s) + NH4Cl heat
NH4NO2(s) heat
NaCl + NH4NO2
N2(g) + 2H2O(l)
(iii) Ammonium nitrite is unstable and explosive
(e) State;
(i) How nitrogen is manufactured
(ii)Three uses of nitrogen.
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Answer
(i)
Nitrogen is manufactured by fractional distillation of air.
(ii)
Nitrogen is used for the manufacture of ammonia
 Liquid nitrogen is used as a cold and inert medium for storage e.g. semen for artificial insemination
 Nitrogen is used as a rocket fuel.
2. (a) The table below refers to the oxides of nitrogen. Complete it correctly under the given headings.
Oxide
For
Preparation
Chemical properties
I
II Action with glowing
mula
(i)
(ii)
Nitrous
N2O
oxide or
splint?
Nitrogen
III Action with hot
(I) Oxide
copper?
Nitric
IV
Heating
V. Action with oxygen?
oxide or
copper in
VI. Action with strongly
Nitrogen
moderately
burning phosphorus?
(II)
conc. Nitric
Extinguishes a burning
Oxide
acid HNO3.
candle and burning
3Cu(s) +
sulphur
8HNO3(aq)
VII. Action with Iron II
3Cu(NO3)2(aq) sulphate?
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+ 2NO(g) +
4H2O(l)
(iii
Nitrogen
VIII
Heating
- Supports
)
dioxide
Pb(NO3)2 or
combustion of
or
reacting
burning phosphorus
Nitrogen
conc. HNO3
(IV)
with copper
burning candle and
oxide
by heating
sulphur
- Extinguishes
- Yellow liquid at
low temperatures is
N2O4.
- On warming
decomposes
N2O4
2NO2(g)
IX
Equilibrium
between
1400C
and 6200?
Answer
I.
Heating a mixture of (NH4)2 SO4 and KNO3.
II.
It rekindles a glowing splint
III.
It is reduced by hot copper to form nitrogen
N2O(g) + Cu(s)
CuO(s) + N2(g)
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IV.
NO
V.
Reacts with oxygen to form a red brown gas NO2.
2NO(g) + O2(g)
2NO2(g)
VI.
Strongly burning phosphorus continues to burn in nitrogen (II) Oxide
VII.
It forms a brown addition product, nitroso iron (II) sulphate.
FeSO4(aq) + NO(g)
VIII.
NO2
IX.
2NO2
FeSO4.NO(s)
2NO(g) + O2(g)
(b) Below is a table of oxides of nitrogen and their physical properties and tests. Complete the table.
Oxide
Physical
Tests
properties
(i)
Nitrous
oxide N2O
I. Colour?
Sweet sickly
smell
- It rekindles a
glowing splint.
- A test tube filled
II. Solubility in
with gas and
water and pH
inverted over
of solution?
water, removed
III. Action with
nitric oxide?
and shaken
repeatedly soon
dissolves
completely.
(ii)
Nitric
IV.
Colour?
- Forms a dark
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Oxide,
- almost insoluble
brown solid with
NO
and neutral to
Iron (II) sulphate
(iii
Nitrogen
)
dioxide
litmus paper.
V. Effect of air?
VI. Colour?
VII. Colour?
- Condenses
NO2
into liquid at
VIII. Effect on wet
210C
litmus paper?
- Irritating
X. Test tube of gas
smell and
poisonous
inverted over
water?
IX. Solubility in
water and pH
of
solution?
Answer
I.
N2O2(g) is colourless
II.
Fairly soluble in water to give a neutral solution
III.
N2O(g) has no reaction with nitric oxide.
IV.
NO(g) forms a red brown gas in air
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V.
Makesi mangi
NO(g) is a colourless gas
2NO(g) + O2(g)
2NO2(g)
VI.
NO2(g) is a red brown gas
VII.
NO2(g) is a red brown gas
VIII.
NO2(g) turns blue litmus paper to red but has no effect on red litmus paper.
IX.
Gas is very soluble and pH of solution = 2
X.
Gas filled into test tube and inverted over water readily dissolves and test tube fills with water.
Q3.
Below is a diagram of arrangement of apparatus used to prepare and collect gas X. Use it to answer the questions that follow.
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(a) Name
(i) Gas X
(i) Liquid Y
(ii) Solid Z
Answer
(i)
Ammonia
(ii)
Water
(iii)
Calcium Oxide
(b) What is the purpose of solid Z?
(ii) Explain with equations why each of the following cannot serve the same purpose as Z in the preparation of gas X.
I.
Calcium chloride
II.
Sulphuric acid
Answer
(i)
Solid Z is used to dry the gas
(ii)
I. Calcium chloride reacts with gas X to form a complex salt
CaCl2 + 8NH3
CaCl2.8NH3
II. Sulphuric acid also reacts with the basic ammonia gas;
2NH3(g) + H2SO4(l)
(NH4)2SO4(s)
(c) (i) Why is the round bottomed flask tilted downwards
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(ii) What method of gas collection is demonstrated in this experiment and what does it tell you about gas X?
(iii) Why is gas X not collected over water?
(d) Write equation for reaction producing gas X.
Answer.
Ca(OH)2(s) + 2NH4Cl(s)
2NH3(g) + H2O(l) + 6Cl3(s)
Answer
To prevent condensed vapour running back into hot flask as this may crack it.
(i)
Upward delivery of gas over air. The method shows the gas is less dense than air.
(ii)
Gas X, NH3 is very soluble in water.
(e) (i) Illustrate on diagram how an aqueous solution of gas X can be prepared.
(ii) Gas X is the most soluble gas in water. An aqueous solution of gas X is
15M. Calculate this concentration in grams per litre (X=17)
Answer
(e) (i) gas X
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Funnel
Trough
Water
(ii) Mass = moles x relative mass
= (15x17)
= 255g/litre
(f) Aqueous ammonia forms a weakly basic solution. Write an equation to
illustrate the behaviour of ammonia in water.
Answer
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
(g)
(i) Under what conditions does gas X react in Air and Oxygen
(ii) Write equation of reactions in (i) above.
Answer
(i) I. Gas X (ammonia) does not react in air ordinarily but it will react in presence of heated catalyst (Pt)
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II. Gas X (ammonia) reacts with oxygen both in the presence and absence of catalyst.
(ii)In the presence of catalyst both air and oxygen react to give same product
4NH3(g) + 5O2(g)
Pt
4NO(g) + 6H2O(l)
In the absence of catalyst only pure oxygen reacts with X (ammonia)
4NH3(g) + 3O2(g)
2N2(g) + 6H2O(l)
(h) State the observations and write equations of reactions when;
(i) Gas X is passed over heated copper II Oxide
(ii)A glass rod dipped into concentrated hydrochloric is held near a bottle with concentrated ammonia solution.
Answer
(i)
The black copper II Oxide changes to red brown solid. Colourless liquid condenses and gas produced puts off a burning splint
and it is neutral to litmus paper.
Equation:
(ii)
NH3(g) + CuO(s)
Cu(s) + H2O(l) + N2(g)
Concentrated white fumes of ammonium chloride are formed.
Equation:
NH3(g) + HCl(g)
NH4Cl(s)
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Q4.
Makesi mangi
Below is a flow chart for Haber process. Study it and answer the questions that follow.
Gas P
Purification of
gas P Compressor
Gases P and Q and Q
Gas Q
recycling
pump
Chamber Y
Ammonia
condenser
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Catalytic
converter
Liquid NH3 to storage
NH3 + H2 + N2
(a) (i) In the purification stage, soda lime will have two purposes, State them
(ii) Name three impurities found in the mixture of P and Q before purification.
(iii) Why is it important to remove impurities.
Answer
(i)
To remove both water vapour and carbon dioxide.
(ii)
Carbon monoxide, dust particles and Noble gases
(iii)
Impurities poison catalyst and make it ineffective.
(b) State the sources and identify;
(i)
Gas P
(ii)
Gas Q
Answer
(i)
Air to obtain nitrogen, gas P
(ii)
Hydrocarbons or water to obtain hydrogen gas, Q
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(c) i) What are the conditions for the manufacture of ammonia.
(ii) Why is it necessary to use;
I.
High pressure
II.
High temperature
Answer
Temperature of 4500C
(i)
Pressure of 250 atmospheres, iron catalyst.
(ii)
I. High pressures favour higher yield of ammonia
II. High temperature speeds up the rate of reaction.
(d) (i) Write equation of reaction between gas P and gas Q to produce ammonia.
(ii) What two things are recycled?
(iii) How is ammonia separated from the unreacted gases?
Answer
(i)
N2(g) + 3H2(g)
2NH3(g)
(ii)
Unreacted gases and heat from chamber Y
(iii)
Ammonia is separated from unreacted gases by cooling to condense into liquid.
(e) (i) Name chamber Y
(ii) What is the purpose of chamber Y? Explain.
Answer
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(i)
Heat exchanger
(ii)
To transfer heat from the gases from the catalytic converter to unreacted gases. This ensures that energy from the catalytic
converter is not wasted but transferred to pre-heat the reactant gases.
Q5. a) i) State four uses of ammonia
(ii)
Write equation of reaction between ammonia and
I.
Phosphorus (V) acid solution
II.
Nitric acid solution.
Answer
(i)
Uses of ammonia are:
- Used in laundries to remove grease
- Manufacture of ammonium nitrate, used in manufacture of explosives and as a fertilizer
- Manufacture of nitric acid
- Manufacture of sodium carbonate
(ii)
I.
3NH3(g) + H3PO4(aq)
(NH4)3 PO4(aq)
II.
HNO3(aq) + NH3(aq)
(NH4 NO4(aq)
(b) Below is an experimental arrangement used to find out what elements
ammonia contains.
Black copper II Oxide
Combustion tube
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Bubbles of
Gas Y
Heat
Quicklime
Pumice stone
Ice
Beaker
Liquid X
(i)
What is the purpose of ;
I.
Pumice stone
II. Quicklime and what is its chemical name?
III. Ice
(ii)
Name
I.
Liquid X
II. Gas Y, and what is the test for each.
(iii)
Write equation of reaction between ammonia gas and copper II Oxide to explain observation in combustion tube.
Answer
(i) I. Pumice stone will enable small bubbles to form and boiling (although we
should avoid this) will be smooth.
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II. Quicklime is calcium oxide and it is used to dry ammonia gas
III. Ice is to cool any water vapour and condense it.
(ii) I. Liquid X is water II. Gas Y is nitrogen
Liquid X turns white anhydrous copper sulphate to blue. Gas Y puts off a burning splint but has no effect on limewater nor
litmus paper.
(iii) 3CuO(s) + 2NH3(g)
N2(g) + 3H2O(l) + 3Cu(s)
A red brown residue left in combustion tube.
(c) Study the table below , complete and answer questions that follow;
Observations and explaining ionic equations on adding:
Ion in (aq.)
A few drops of
A few drops
Excess
Excess
Solution
NH3(aq)
of NaOH(aq)
NH3(aq)
NaOH(aq)
A (i) Mg2+
White
I
precipitate
2OH-(aq) +
Mg2+(aq)
Mg(OH)2(s)
(ii) Ca2+
White
precipitate
2OH(aq)- +
Ca(OH)2(s)
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B (i) Al3+
Makesi mangi
White
White
precipitate
precipitate
3OH(aq)- +
II
dissolves
Al3+(aq)
Al(OH)3(s) +
Al(OH)3(s)
3OHAl(OH)4(aq)-
(ii) Zn2+
White
III
White
precipitate
precipitate
2OH(aq)- + Zn2+
dissolves
Zn(OH)2(s)
Zn(OH)2(s)
IV
+ 4NH3(aq)
Zn(NH3)42
+(aq) +
(iii) Pb2+
White
2OH-(aq)
precipitate
Pb2+(aq) +
V
C (I) Fe2+
2OH-(aq)
VI
Pb(OH)2(s)
Dirty green
VII
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precipitate
Fe2+(aq) +
(ii) Fe3+
2OH(aq)Fe(OH)2(s)
Redish
(iii) Cu2+
VIII
brown
precipitate IX
Royal
XI
blue
solution
Blue
X1I
precipitate
X
(i)
Complete the table for the spaces numbered I-XII
(ii)
What is common about the ions in each of the divisions A, B and C.
(iii)
Explain why I. Ca2+(aq) ions do not form a precipitate in NH3(aq)
II. The complex ion Al(OH)4-(aq) does not form in NH3(aq). Use any relevant equation.
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(iv)
Makesi mangi
Give the formula of;
I.
Precipitate in B(II) and B(V)
II. The complex ion in B(IV) and C(XII)
Answer
(i)
I.
2OH-(aq) + Mg2+(aq)
II.
3OH(aq)- +Al(aq)3+
III.
White precipitate, Zn2+(aq) + 2OH(aq)-
IV.
Zn(OH)2(s) + 2OH(aq)-
Mg(OH)2(s) White precipitate
Al(OH)3(s) White precipitate
Zn(OH)2(s)
Zn(OH)42-(aq) White precipitate
dissolves
V.
White precipitate, Pb(aq)2 + 2OH(aq)-
Pb(OH)42-(aq)
VI.
White precipitate dissolves, Pb(OH)2(s) + 2OH-(aq)
Pb(OH)42-(aq)
VII. Dirty green precipitate, Fe2+(aq) + 2OH(aq)-
Fe(OH)2(s)
VIII. Reddish brown precipitate, Fe3+(aq) + 3OH-(aq)
Fe(OH)3(s)
IX.
Fe3+(aq) + 3OH-(aq)
IX.
Cu2+(aq) + 2OH-(aq)
X.
Blue precipitate, Cu2+(aq) + 2OH(aq)
XI.
Cu(OH)2(s) + 4NH3(aq)
Fe(OH)3(s)
Cu(OH)2(s)
Cu(OH)2(s)
Cu(NH3)42+(aq) + 2OH-(aq)
Blue precipitate dissolves
(ii)
In A all the ions form a white precipitate in excess NaOH(aq)
In B all the ions form a white precipitate soluble in excess NaOH(aq)
In C all the ions form coloured hydroxide precipitates
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(iii)
Makesi mangi
I. Calcium hydroxide is a stronger base than ammonia so ammonia cannot provide it with enough OH-(aq) to form a
precipitate.
II. Ammonia is a weak base.
NH4+(aq) + OH-(aq)
NH3(aq) + H2O(l)
It is only partially dissociated into ions and the OH- ions formed are few to form the al(oh)-4 complex ion or other hydroxide
complexes. Most ammonia exists as dissolved molecules, NH3(aq). Where possible complexes form with the molecules as in
Zn(NH3)42+ and
(iv)
Cu(NH3)22+
I. Precipitate in B(II) = Al(OH)3
B(V) = Pb(OH)2
II. The complex ion B(IV) = Zn(OH)42-(aq)
B(XII) = Cu(NH3)42+(aq)
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Q6.
Makesi mangi
The diagram below illustrates three stages into which the manufacture of nitric acid can be split.
water
Heated
A and B
C
Platinum
cooling
Gauze
Tower
and
B
300C
Nitric
acid
Ammonia and Air
STAGE I
STAGE II
STAGE III
(a) Name the gases A, B, and C.
Answer
A is nitric oxide, NO(g)
B is Air
C is nitrogen dioxide, NO2
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(b) Write equation of reaction at
(i)
Stage I
(ii)
Stage II
(iii)
Stage III
Answer
(i)
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(l)
(ii)
2NO(g) + O2(g)
(iii)
4NO2(g) + 2H2O(l) + O2(g)
2NO2(g)
4HNO3(l)
(c) What is the purpose of
(i)
Heated platinum gauze?
(ii)
Cooling chamber?
(iii)
Broken bricks?
Answer
(i)
Heated platinum gauze acts as a catalyst
(ii)
To cool mixture of nitric oxide and air to react and form nitrogen dioxide
(iii)
To provide a large surface area for nitrogen to dissolve in water.
(d) What temperature is required for reaction between ammonia and air and how is this temperature attained.
Answer
The temperature is 700-9000C. This temperature is achieved by first heating platinum gauze electrically but once reaction starts, heat
from reaction maintains the temperature as required.
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(e) Suggest appropriate descriptive terms than may be used to identify stages I, II and III.
Answer
Stage I – Catalytic oxidation – of ammonia
Stage II – Cooling of nitric oxide/air mixture.
Stage III – Absorption stage.
(f) To a pale green solution Y, Dilute sodium hydroxide solution was added. A dirty green precipitate formed. On adding
concentrated nitric acid the green solution Y turned yellow. On adding sodium hydroxide solution a brown precipitate now
formed. Explain observations using equations identifying the ion in Y.
Answer
The ion in Y is Fe2+
Fe2+(aq) + 2OH-(aq)
Fe(OH)2(s) – dirty green precipitate
On adding conc. Nitric acid, Fe2+ was oxidised to Fe3+.
Fe2+(aq) + NO-3(aq)
Fe3+(aq) + NO2 + O2-(aq)
(f) Write equation of reaction between conc. Nitric acid and;
(i) Iron II sulphate in presence of dilute sulphuric acid.
(ii) Sulphur
Answer
(i) 2HNO3(aq) + 6FeSO4(aq) + 3H2SO4(aq)
(ii) S(s) + 4HNO3(l)
3Fe2(SO4)(aq) + 4H2O(l) + 2NO(g)
SO2(g) + 4NO2(g) + 2H2O(l)
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(h) State two uses of nitric acid.
Answer.
- Manufacture of nitrate fertilisers e.g. NaNO3
- Manufacture of ammonium nitrate, both a fertiliser and an explosive.
(i)
A mixture of compounds P and Q was subjected to the following
tests.
A: Appearance: Light green crystalline solid.
B: To a solution of mixture was added a few drops of dil sodium hydroxide solution. A dirty green precipitate formed.
C. Addition of potassium hexacyanoferrate (III) to solution gave a dark blue precipitate. A red brown gas also produced. Gas
has pungent smell and changes wet blue litmus to red.
D. Addition of aq. FeCl3 to solution gave a light green solution and a red brown gas
E. To the mixture in test tube was added a small quantity of conc. H2SO4. Oily drops on side of test tube seen on heating.
Brown fumes also form.
F. Add a few drops of conc. Sulphuric to solution in test tubes Brown solution formed and red brown gas produced.
I. In each of the observations write your inference.
II. Deduce the cation and anions present in mixture.
Answer.
I.
A - Mixture contain Fe2+ or Cu2+ salts.
B - Fe2+ ions likely
C - Fe2+ ions confirmed. NO2- ions also likely
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D – Reducing ions e.g NO2- ion present
E – NO3- ions and NO2- ions likely
F - Fe2+ and NO2- ions confirmed
II.
Fe2+ (Cation)
NO2- and NO3- (anions).
Q7.
The table below is a record of observations when of W, X, Y Z and R, not true element symbols heated in a test
tubes in turns.
Nitrate
W
Observations
On gently heating steamy fumes condensing in cooler
parts to produce a colourless liquid. On stronger
heating crackling sound is heard and gas produced
re-lights a glowing splint. Red brown gas produced
turns wet blue litmus red and a silvery white liquid is
condensed.
X
Melts into a colourless liquid. White fumes are
produced. Colourless liquid condenses in cool parts
and gas re-kindles glowing splint. No residue is left.
Y
On strong heating a crackling sound heard. Gas
produced re-lights a glowing splint . Wet blue litmus
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change to red and a red brown gas produced. Residue
left is yellow but it is brown when hot.
Z
White crystals melt into a colourless liquid. Gas
produced re-lights a glowing splint and a pale yellow
residue is formed when solid is very strongly heated.
Q
A green solid on gently heating steamy fumes and
colourless liquid condenses in cooler parts of test
tube. On stronger heating crackling sound is heard
and gas formed re-lights a glowing splint. Red brown
gas formed and a black residue is formed.
(a) Suggest the identity of the nitrate X.
Answer
Ammonium nitrate, NH4 NO3
(b) Suggest the identity of the nitrates of Y and Q.
Answer
Y = Pb(NO3)2
Q = Cu(NO3)2
(c) List the metals of the nitrates W, Y, Z and Q in their decreasing order of reactivity.
Answer
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ZYQW
(d) Suggest the identity of W and Z
Answer
W = Mercury II nitrate, Hg(NO3)2
Z = Sodium (or potassium) nitrate, NaNO3 or KNO3
(e) Identify the colourless liquid condensed on heating Q and hence name the nitrate of Q
Answer
Water, Hydrated copper II nitrate
(f) Write a balanced equation for the decomposition of hydrated copper II nitrate, Cu(NO3)2. 3H2O.
Answer
2Cu(NO3)2.3H2O(l)
2CuO(s) + 4NO2(g) + O2(g) + 6H2O(l)
(g) Below is an incomplete description of the preparation of ammonium salt.
Apparatus.
Dilute nitric acid
Q
R
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P
Dilute aqueous ammonia with 3 drops of methyl orange.
25.0cm3 of dilute aqueous ammonia was added into container P.
Results – Burette readings.
Final
Rough
1st
2nd
3rd
25.0
24.5
48.8
24.5
0.0
0.0
24.4
0.0
25.0
24.5
24.4
24.5
(cm3)
Initial
(cm3)
Differenc
e (cm3)
(i)
From the table of results what was the volume of acid required to neutralise 25.0cm3 of ammonia solution?
(ii)
If the colour of methyl orange in alkali is yellow and in acids it is red-orange. What is the colour of methyl orange.
I.
At end-point of reaction
II.
In sodium hydroxide
III.
In dilute nitric acid.
(iii)
If the dilute aqueous ammonia was exactly 2M, calculate the molarity of dilute nitric acid.
(iv)
Calculate maximum mass of ammonium nitrate formed.
(H = 1.0, N = 14.0, O = 16.0)
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(v)
Makesi mangi
What is the error in the volume of acid used?
B. A clean container was now taken and 25.0cm3 of alkali added. No indicator was added. Same volume of acid from apparatus Q
was then run into the container. This mixture was used to obtain salt.
(i)
Name the process used to obtain salt form solution.
(ii)
Name all the apparatus required
(iii)
A water bath should be used. Explain
(iv)
What colour of crystals is obtained.
Answer
A. (i) Volume = 24.5cm3
(ii) I. Pink at end-point
II. Yellow in NH3(aq) – (NH4 OH(aq))
III.Red-Orange in nitric acid.
(iii) Moles of NH3(aq)
Equation:
Or
= (2/1000 x 25.0)
NH3(aq) + HNO3(aq)
= 0.005mole
NH4 NO3(aq)
NH4 OH(aq) + HNO3(aq)
NH4 NO3(aq) + H2O(l)
Mole ratio is 1:1
Mole of acid = 0.005
Molarity of HNO3
(i)
= (0.05/24.5 x 1000)
Moles NH3(aq) = moles
But
NH4 NO3
= 2.04M
NH4 NO3 (ratio is 1:1) from equation in (iii) = 0.05mole
= (14+1x4+14+16x3)
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Makesi mangi
= 80
Mass
= (0.05x80)
= 4.0g
(ii)
Error in vol. Of acid (titration)
= 1drop
= 0.05cm3
i.e. colour change at end-point is caused by 1 drop.
B.
Answer
(i)
Evaporation
(ii)
Evaporation dish or crucible, bunsen burner, beaker, tripod stand and wire gauze.
(iii)
Ammonium nitrate decomposes on heating therefore water bath ensures temperature is kept low so that little decomposition
occurs.
(iv)
Ammonium nitrate crystals are white.
PRACTICE QUESTIONS
(a)
State two physical and two chemical properties of nitrogen.
(b)
There are two methods commonly used in preparing nitrogen in the laboratory. State these two methods giving equation for
important or major reactions involved.
(c)
Lithium reacts with nitrogen at room temperature to produce a solid;
(i) Write equation of reaction
(ii) Write equation of reaction between solid and water.
(iii) Why is this observation unique?
(d)
Name the gases of the air giving percentage composition
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(e)
Makesi mangi
(i) What is the method of manufacture of nitrogen
(ii) Briefly state the manufacture of nitrogen.
(iii) State three uses of nitrogen.
(f)
Use the group and period of nitrogen to identify its position in the Periodic Table.
Q2. (a) State one use for each of the oxides
(i)
Nitrogen (I) Oxide (nitrous oxide)
(ii)
Nitrogen (II) Oxide (nitric oxide)
(iii)
Nitrogen (IV) Oxide (nitrogen dioxide)
(b) The diagram below shows a method of preparing an oxide of nitrogen.
Mixture of X and
Potassium
gas Z
nitrate
Trough
Water
Heat
(i) To produce the oxide Z, a mixture of solids X and potassium nitrate have been used instead of the single substance ammonium
nitrate. Name solid X.
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(ii) Why is it that ammonium nitrate is not used but a mixture of two salts X and Potassium nitrate used to produce gas Z?
(iii) Name gas Z.
(iv) Write equations for reactions producing gas.
(c)
(i) Explain how nitrogen IV Oxide, NO2 is produced and collected from lead II nitrate.
(ii) In what form is nitrogen IV oxide collected. Give the actual molecular formula of the oxide collected.
(d) The action of moderately concentrated nitric acid on copper produces nitrogen II Oxide, salt and water.
Write equation of reaction.
(e) (i) Name three Oxides of nitrogen and state two physical properties for each.
(ii) Complete the following reactions of oxides of nitrogen.
red heat
I.
N2O(g)
II.
P4(s) + N2O(g)
+
a gas and cloud of white
smoke
r.t.
III.
NO(g) + O2(g)
IV.
NO(g) + P4(s) (Strongly burning substance)
V.
FeSO4(aq) + NO(g)
VI.
NO2
VII.
NO2(g) + P4(s) (Strongly burning substance)
VIII.
NO2(g) + H2O(l)
1400 – 6200C
+
+
+
+
Q3. Simple test tube tests given in the table below were done on unknown compound X. Complete the inferences and answer
questions that follow.
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Test
Observation
Inference
A. Appearance of X
Solid is crystalline
Fe3+ ion likely
and has a red brown
be present.
colour.
B. Substance X was heated
Colourless liquid
NH4+, Fe3+ and
gently at first and then
condenses in cool
ion that
strongly.
parts of test tube.
decomposes
Gas produced turns
to acidic gas
red wet litmus to
e.g.
blue. On further
SO32-, SO42-,
heating a gas which
etc.
turns wet blue litmus
Likely to be
to red is produced. A
present
C. Dil. Sodium hydroxide
solution added in excess
red brown powder
residue left.
and warmed
Red brown
precipitate.
I
Gas that turns wet
red litmus produced
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D. A few drops of barium
chloride solution was
A white precipitate
Makesi mangi
II
formed
added followed by a few
drops of dilute sulphuric
acid.
E. Prepare solution X and
then to1cm3 of solution add
Solution becomes
1cm3 of dilute sulphuric
decolourised
acid and some iron filings
A dirty green
and warm.
precipitate formed
III
Allow all the hydrogen gas
to bubble through solution
until no further change. Add
a few drops dil. NaOH.
(a) (i) Complete the inferences I – III in table
(ii) Write formula for all the ions detected and any molecule present
(b) (i) What is the action of iron in test E?
(ii) Why is iron used instead of Zinc in test E?
(iii) Explain the production of hydrogen in test E by using an equation.
(c) The compound X is thought to have the formula (NH4)x Fey (SO4)4.2
i) What type of salt is this
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ii) Suggest possible values of x and y
d) (i) Write ionic equation for reaction producing alkaline gas in test C.
(ii) Give an application of this reaction in the laboratory and explain using full
chemical equation.
(e) When orange ammonium dichromate salt, (NH4)2 Cr2 O7
is heated a green oxide (Cr2O3), Oxygen and two other products are formed.
Write an equation for the decomposition.
Q4. Ammonia is manufactured from nitrogen and hydrogen by the Haber process. The equation for the exothermic reaction
is;
N2 + 3H2
2NH3
(a) What type of reaction is this represented by the equation shown?
(b) What impurities are usually removed from reactant gases and why?
(c) (i) What are the conditions for the manufacture of ammonia?
(iii) State exactly how each condition is achieved.
(d) Explain how ammonia is separated from unreacted gases and the importance of recycling of;
(i) Gases
(ii) Heat
(e)
(i)
Name and give the formula of two commercially important
compounds of ammonia.
(ii) State the importance of the compounds named.
(f)
(i)
Ammonia can be oxidised to nitrogen oxide or nitrogen in
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presence of oxygen also forming water in each case.
State conditions and write equations in which the two different
products are formed.
(ii) Complete the following reactions of ammonia and; Balance them:
I. CuO(s) + NH3(g)
+
+
II. Cl2(g) + NH3(g)
+
+
III. NH3(g) + HCl(g)
+
+
IV. NH3(aq) + H2O(l)
+
+
(g) In the reaction between copper II oxide and ammonia, 1dm3 of the gas measured at room temperature was reacted.
(i) What was the mass of water formed
(ii) Volume of nitrogen gas formed at room temperature and pressure
(H = 1.0, O = 16.0, 1 mole of gas at r.t.p and pressure conditions
= 24dm3)
Q5(a) Ammonia gas dissolves in water to give a weak an alkaline solution.
(i)
Write an equation to show this.
(ii)
I. Starting with ammonium chloride and calcium hydroxide write
equation for reaction.
II. Name drying agent for ammonia
III. What method of gas collection is used?
IV. Draw diagram to show how a solution of ammonia can be prepared from gas so prepared.
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(iii) The clip of a rubber tube at the mouth of a round bottomed flask held upside down was opened under water having a few drops
of phenophthalein. A red fountain formed. Explain this observation.
(iv) Write the formula of all the ions and molecules found in solution of ammonia
(v)
State the two covalent bonds in an
NH4+ ion
(vi) What is observed when solid calcium oxide is added to a solution of ammonia.
(b)
The following reactions can occur;
(i)
2Cu(s) + 2NO(g)
(ii)
FeSO4(aq) + NO(g)
(iii) 2NO(g) + O2(g)
2CuO(s) + N2(g)
FeSO4.NO(aq)
2NO2(g)
Under what conditions do each of the reactions occur.
(c) When a mixture of dilute sulphuric acid and potassium nitrate are heated gently a pale yellow liquid gradually distills over.
(i)
What was the pale yellow liquid
(ii)
Write equation of reaction between dilute sulphuric acid and potassium nitrate
(iii)
Why was the liquid pale yellow
(iv)
Nitric acid is prepared in an all glass apparatus. Explain
(v)
What is the maximum mass of the pale yellow liquid obtained by reacting 15g of potassium nitrate in sufficient dilute nitric
acid.
(d) The pale yellow liquid in (c) above was used to dissolve 1.0g of copper in a fume cupboard. There was a vigorous reaction;
producing a blue solution and a red brown gas. The solution was heated until no more brown fumes to produce a black powder.
(i) Write equation of reaction between copper and the pale yellow liquid.
(ii) What mass of black solid was produced
(Cu = 64, O = 16, K = 39, N = 14, H = 1)
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Q6(a) (i) State three uses of ammonia
(ii) Write equation of reaction between ammonia and
I.
Sulphuric acid, H2SO4
II.
Phosphoric acid, H3PO4
III.
Nitric acid, HNO3
(b) Aqueous ammonia, like aqueous sodium hydroxide solution will form a precipitate with cations like Fe2+, Fe3+, Cu2+, Al3+, Mg2+,
Pb2+ and Zn2+. It does not form a precipitate with Ca2+
(i) Write formula for ion responsible for formation of precipitates
(ii) Write equation for the brown precipitate with Fe3+ ion salt solutions
(iii) The precipitates with Zn2+ and Cu2+ will dissolve in excess NH3(aq)
I.
What is seen when a few drops of NH3(aq) is added in each case?
II. What is seen when excess NH3(aq) is added in each case?
III. Write equations for the reactions producing the two clear solutions.
(iv) Explain why Ca2+ ions are not precipitated by NH2(aq).
(v)
I. Write formula of the three ions which form white precipitate soluble in excess. NaOH(aq) solution.
II. Write formula of ions formed when the clear solutions form for each of the three ions.
III. How would you distinguish presence of Pb2+ ions in solution from that of Al3+.
(c) Nitric acid is prepared in the laboratory by heating conc. Sulphuric
acid with potassium nitrate in a glass retort. The receptacle of the glass
retort leads to a round bottomed flask with cold water running over it.
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Brown fumes are seen in the retort and oily droplets form in the flask.
(i) Compare boiling point of nitric acid and that of sulphuric acid.
(ii) What are the brown fumes inside the glass retort. Write equation to explain.
(iii) What is the purpose of cold water?
(iv) Write equation for the production of the oily droplets.
(v) An all glass apparatus is used. Explain.
d) Nitric acid reacts both as a typical acid and as an oxidising agent.
(i) Give three properties that show nitric acid as a typical acid and illustrate with equations where possible.
(ii) Explain these two reactions of nitric acid.
4Zn(s) + 10HNO3(aq)
4Zn(NO3)2 + 5H2O(l) + N2O(g)
Mg(s) + 2HNO3(aq)
Mg(NO3)aq + H2(g)
(Very dilute)
(iii)
(iv)
Complete the following reactions.
I.
H2S(g) + 2NHO3(aq)
+
+
NO
II.
P(s) + 5HNO3(aq)
+
+
NO2
III.
C(s) + 4HNO3(aq)
CO2(g) + 2H2O +
IV.
S + 4HNO3(aq)
V.
HNO3(aq) + FeSO4(aq) + H2SO4
Fe2(SO4)3(aq) +
VI.
HNO3(aq) (Conc) + Cu(s)
+
H2SO4 +
+
NO2
+
What is unique about nitric acid on the reactions in d(iii) above.
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(e) State three uses of nitric acid.
CHAPTER IV
Sulphur and compounds
Solved problems
1. The diagram below illustrates the Frasch process for extraction of sulphur.
Hot air (15atm)
Product X and two other
substances
Hot water,
1700, 10atm
Three concentric pipes
200 metres or
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more below the
ground
sulphur deposits
sulphur deposits
12345
(a) (i) Name X and the other two substances coming out of pipe B at end.
(ii) In what state is product X? Explain
Answer
(i) X = Sulphur and the other two substances are water and air.
(ii) In molten state. Sulphur melts below 1700C.
(b)What is the purpose of;
(i) Super-heated water?
(ii) Hot air?
Answer
(i) To melt the sulphur
(ii) To force sulphur up the pipes under pressure.
(c) How is superheated water at 1700C obtained.
Answer: Boil the water under higher pressure conditions.
(d) Identify the arrows 1, 2, 3, 4 and 5 with the correct substance passing through it.
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Answer:
1 = Hot water,
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2 = product X and other substances
3 = Hot air, 4 = Product X and other substances
5 = Hot water
(e) Sulphur may also be extracted from natural gas;
(i) How is hydrogen sulphide separated from a mixture with natural gas, CH4
(ii) In partial oxidation of hydrogen sulphide, sulphur dioxide and water are first formed. The sulphur then oxidises more
hydrogen sulphide to sulphur.
I. Write equations of reactions
II. What fraction of hydrogen sulphide needs to be converted to sulphur dioxide in the partial oxidation.
Answer:
(i) The hydrogen sulphide is absorbed in special solvents – alkaline solution and then mineral acid added to liberate H2S.
(ii) I. H2S + O2(g)
2H2S + SO2(g)
SO2(g) + H2O(l)
3S + 2H2O
II. From equations above only one third of the hydrogen sulphide needs to be oxidised in air to sulphur dioxide.
Q2.(a) Sulphur melts at 1130C to a clear amber liquid, which flows easily. It darkens on further heating and at about 1600C it becomes
reddish brown and viscous. It boils at 4440C forming a brownish gas before which it becomes mobile again.
(i) Explain the changes observed
(ii) The sulphur boils as S8 molecules but which at 10000C may form diatomic molecules, S2 and at higher temperatures still,
molecules are monoatomic.
I. What name is given to the three gaseous forms of sulphur
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– S8, S2 and S?
II. Write balanced equations for the changes
III.
Sulphur displays allotropy in all three physical states. Explain.
Answer.
(i) Sulphur melts as S8 rings and these can easily move. These rings break into S8 chains which soon join to form long chains. These
entangle and make the sulphur viscous. On further heating the liquid becomes mobile again and black. The chains break and
become shorter.
(ii)
I.
Allotrophy
II. S8(s) (ring)
S8(l) (chain) (1600C)
100S8(l) (chain)
(iii)
100S8(l) (4440C)
S8(g)
4S2(g) (10000C)
S2(g)
2S(g)
>> 10000C.
In solid state we know sulphur has three allotropes mainly monochronic, rhombic sulphur and plastic sulphur. In the liquid
state we also have sulphur in allotropic forms e.g. molten runny sulphur with S8 rings, molten sulphur with sulphur chains.
In the gaseous state we also have S8(g) in forms of rings below 10000C but which break down to S2 and even to monoatomic
S(g) molecules at higher temperatures.
(a)
State three differences between monochronic and rhombic sulphur.
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Answer:
Rhombic
Monoclinic
Octahedral crystals
Needle like crystals
Stable below 960C
Stable above 960C
Melts at 1150C
Melts at 1200C
Bright yellow
Pale yellow
Density higher (2.06cm-3)
Density lower (1.98gcm-3)
(b)
The scheme below summarises reactions starting from sulphur.
A
B
Fe,
Heat
O2, heat
Products
C, D
and E
Conc. HNO3
S White hot, Carbon
Liquid F
heat
Reagent C and
heat
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D+B
Name products A, B, C, D, E and F
Answer:
A = FeS, B = SO2, C = H2SO4.
D = H2O, E = NO2, F = CS2
(c) State three uses of sulphur.
Answer: Sulphur is used in manufacture of sulphuric acid;
- making of matches
- Manufacture of insecticides, fungicides and drugs.
Q3. The arrangement in diagram below is a set-up used to prepare and collect
dry sulphur dioxide.
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Conc. H2SO4
Card cover
Substance X
SO2
Heat
Liquid M
(a)
(i) Name substance X
(ii) State your observations
(iii) Name liquid M and state its purpose
(iv) Name the method of gas collection.
Answer:
(a)
(i) Copper
(ii) Bubbles of gas, gas has pungent smell, and wet litmus turns red, blue solution formed.
(iii) Sulphuric acid to dry sulphur dioxide.
(iv) Downward delivery over air.
(b) (i) State two reasons why heating is necessary
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(ii) When sodium sulphite is used to prepare sulphur dioxide what other substance is required.
Answer (b)(i).
(b)
Conc. Sulphuric acid oxidises copper when heated and sulphur dioxide is also a very soluble gas and heating is required to
release it from solution.
(ii) Dilute acid e.g. hydrochloric or sulphuric acid;
(c) (i) Name the product formed when sulphur dioxide dissolves in water and write equation for its formula.
(ii) What is seen when sulphur dioxide (in moist conditions) is bubbled through.
I.
Conc. Nitric acid.
II. Potassium manganate VII solution
III. Red bromine solution
IV. Blue or red flower extract solution
(iii) Write equations of reactions in (ii) I-IV above.
(iv) State one advantage and one disadvantage chlorine over sulphur dioxide as a bleaching agent.
Answer
(c)
(i) Sulphurous acid.
H2O(l) + SO2(g)
H2SO3(aq)
(ii) I. Reddish brown nitrogen dioxide is produced and mixture becomes warm.
II. Purple permanganate solution changes to pale pink
III. Red bromine solution decolourises.
IV. It becomes decolourised.
(iii) I. 2HNO3(l) + SO2(g)
H2SO4(l) + 2NO2(g)
II. 2KMnO4(aq) + 5SO2(g) + 2H2O(l)
K2SO4(aq) + 2MnSO4(aq) + 2H2SO4(aq)
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III. SO2(g) + Br2(l) + H2O(l)
H2SO4(aq) + 2HBr(aq)
IV. Dye + SO2 + 2H2O(l)
(Dye + Hydrogen) + H2SO4
(iv) Chlorine bleaches by oxidation while sulphur dioxide bleaches by reduction and this is often reversed by action of oxygen from
the air.
- Wool and silk are usually damaged when bleached by chlorine.
(d) State three uses of sulphur dioxide.
Answer.
- To bleach wool, silk, straw and sponges.
- To manufacture Ca(HSO3)2 used to bleach wood-pulp white in manufacture of paper.
- Preservative to foodstuffs and fruit squashes by reacting with oxygen and preventing oxidation of the liquids.
- Used in contact process for manufacture of sulphuric acid etc.
(e) Burning magnesium continues to burn in sulphur dioxide producing brown fumes and a white powder and yellow deposit left.
When a gas jar of sulphur dioxide is inverted over a gas jar of hydrogen sulphide, yellow solid forms.
(i)
Name the white solid and yellow solid or brown fumes.
(ii)
Explain how each of the reactions occurs.
(iii)
Explain the brown fumes.
(iv)
Write balanced equations of reactions.
Answer:
(i)
White solid is magnesium oxide and yellow solid is sulphur.
(ii)
The intense heat of burning magnesium decomposes sulphur dioxide to sulphur and oxygen. The oxygen is then used to
maintain the burning of magnesium.
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- In the other reaction involving the two gases sulphur dioxide oxidises hydrogen sulphide to sulphur and water.
(iii)
The intense heat of burning magnesium causes sulphur to boil producing brown fumes.
(iv)
Reactions;
2Mg(s) + SO2(g)
2MgO(s) + S(s) and
2H2S(g) + SO2(g)
3S(s) + 2H2O(l)
Q4. Below is a flow-chart showing essential stages in the manufacture of sulphuric acid by contact process.
B
A
SO2
Air Purification
Roasting
Air
Air + SO2
filtration and
drying
Stage 1
Stage II
tower X
Waste gases
conversion
chamber
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Quartz
Stage III
Dilute water
D
Stage IV
To storage
(a)
Name a possible substance represented by A, B, C and D
Answer: A = Iron sulphide
B = Iron III Oxide
C = Sulphur trioxide and other gases like nitrogen and sulphur
dioxide.
D = Oleum.
(b) Name four products that are recycled.
Answer: (b) Sulphur dioxide, sulphuric acid, oxygen and heat.
(c) What are the conditions for manufacture of sulphuric acid?
Answer:
Pressure of 1-2 atmospheres, catalyst – V2O5 or (Pt) and temperatures of 5000C.
(d) Write equations of reactions at;
(i) Stage I
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(ii) Stage III
(iii) Stage IV (two equations)
Answer: (d)
(i) 4FeS(s) + 7O2(g)
2Fe2O3 + 4SO2(g)
(ii) 2SO2(g) + O2(g)
2SO3(g)
(iii) H2SO4(l) + SO3(g)
H2S2O7
H2S2O7(l) + H2O(l)
2H2SO4(l)
(e) What is;
(i) Removed at purification stage?
(ii) The purpose of tower X?
(iii) The appropriate name for stage IV?
Answer: (e)
(i) Water and dust particles.
(ii) It is used as a heat exchanger to transfer heat from hot gases from conversion chamber, to cool incoming gases.
(iii) Stage IV is known as absorption stage.
(e)
Why is it importance to remove impurities?
Answer: (e) Impurities make catalyst ineffective (poison catalyst)
(f) What is the effect of each of the following on (I) yield of sulphuric trioxide and hence the acid at equilibrium (ii) rate of reaction.
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I. Temperature
II. Catalyst
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III. Pressure
IV. Excess air.
(f) (i)
I.
Reaction is exothermic therefore an increase in temperature leads to low yield of acid.
II. Catalyst has no effect on yield.
III. Pressure increase will increase yield of acid as products are formed with reduction in volume.
IV. Excess air will increase yield of acid as a shift in equilibrium occurs to the right producing more sulphur trioxide.
The more expensive sulphur dioxide will be consumed.
(ii)
I.
An increase in temperature increases rate of reaction.
II. Catalyst increases rate of forward reaction
III. Pressure increase will increase rate of forward reaction. An increase in pressure for gases increases the concentration of reactants.
IV. An increase in air will increase concentration and hence an increase in rate of forward reaction.
(g) Explain how environmental pollution occurs in the contact process and how it is controlled?
Answer: The contact process produces acidic gases sulphur dioxide and
sulphur trioxide. These may leak into atmosphere and react with
moisture to produce acids.
SO2(g) + H2O(l)
H2SO3(aq)
SO3(g) + H2O(l)
H2SO4(aq)
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These will mix with rain producing acid rain which corrodes metal and rock surfaces. Pollution by these gases is controlled by passing
any waste gases through alkaline aqueous solution to absorb these acid gases (scrubbing).
Q5. (a) Sulphuric acid is a strong acid. Write an equation to illustrate this by its dissociation in water.
Answer (a):
2H3O+ + SO42-(aq) or
H2SO4(aq) + 2H2O(l)
H2SO4(aq)
2H+(aq) + SO42-(aq)
(b) Concentrated sulphuric acid removes the elements of water from many compounds, Write equation for its reaction with;
(i) Ethanol, C2H5OH at 1600C.
(ii) Glucose, C6H12O6
Answer (b)
(i) C2H5OH(l) + H2SO4(l)
C2H4(g) + (H2SO4(aq) + H2O(l))
(ii) C6H12O6(s) + H2SO4(l)
6C(s) + (6H2O(l) + H2SO4(aq))
(c) What is seen when concentrated sulphuric acid is added to;
(i) Cellulose (wooden splint)?
(ii) Copper sulphate crystals?
Answer (c)
(i) The wooden splint becomes charred to black and bubbles form on surface of splint.
(ii) The blue copper sulphate crystals turn to white powder.
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(d) The ability of concentrated sulphuric acid to remove elements of water from compounds has been made use of in the preparation
of carbon monoxide. Study the diagram below and answer questions that follow.
Conc. Sulphuric acid
CO(g)
Trough
Water
Methanoic acid(HCOOH)
(i)
What is seen in flask?
(ii)
Write equation of reaction.
(iii)
When sodium methanoate (HCOONa) is used in place of methanoic acid reaction occurs in two steps, first producing the acid:
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I.
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What type of reaction is the first step?
II. Write equations for the two reactions.
(iv)
In a similar experiment oxalic acid H2C2O4, a dibasic acid can be used in place of methanoic acid.
I.
Write equations of reaction
II. How are the product gases separated to get pure carbon monoxide.
Answer (d)
(i)
Bubbles of colourless gas in round bottom flask.
(ii)
HCOOH(l) + H2SO4(l) CO(g) + H2SO4(aq) + H2O(l)
(iii)
I. Displacement of a weak acid from its salt.
(iv)
II. HCOONa(s) + H2SO4(l)
HCOOH(l) + NaHSO4(s)
HCOOH(l) + H2SO4(l)
CO(g) + H2O(l) + H2SO4(aq)
I. H2C2O4(l) + H2SO4(l)
CO(g) + CO2(g) + H2O(l) + H2SO4(aq)
II. CO(g) is neutral but CO2(g) is acidic. It is therefore possible to separate CO(g) from mixture by passing it through aqueous
sodium hydroxide solution.
(e)
Sulphuric acid is described as hygroscopic.
(i)
What does this term mean?
(ii)
What use has this property of sulphuric been applied in the laboratory.
Answer (e)
(i) Ability to absorb water from the atmosphere but not to the extent of being dissolved in it.
(ii) Conc. Sulphuric is used as a drying agent.
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Q6. (a) When concentrated sulphuric acid is heated with powdered charcoal, carbon dioxide is formed together with water and sulphur
dioxide. A similar reaction occurs with sulphur.
(i) Write equation for reaction with carbon and sulphur.
(ii) What type of reaction occurs in these two cases. Explain
(iii) Why is it that hydrochloric acid does not react with these two elements.
Answer:
(i) H2SO4(l) + C(s)
2H2SO4(l) + S(s)
SO2(g) + CO2(g) + H2O(l)
3SO2(g) + 2H2O(l)
(ii) Redox reactions e.g. Carbon is oxidised to carbon dioxide (receives oxygen) and sulphuric acid is reduced to sulphur dioxide as it
loses oxygen.
(iii) Hydrochloric acid has no oxidising ability therefore it cannot react with non-metals.
(b) Sulphur dioxide is prepared in the laboratory by heating concentrated sulphuric acid with copper.
(i) Write equation of reaction.
(ii) Why is it that copper does not react with dilute Sulphuric acid.
Answer
(i) Cu(s) + 2H2SO4(aq)
CuSO4(aq) + 2H2O(l) + SO2(g)
(ii) Copper is below hydrogen in reactivity therefore it cannot displace hydrogen from acid. Dilute sulphuric acid also lacks ability to
act as oxidising agent therefore no reaction occurs.
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(c)
(i) What properties of sulphuric acid are made use of in the preparation of other acids?
(ii) Name the chemicals needed to prepare;
I. Nitric acid
II. Hydrogen chloride
(iii) In each case in (ii) above write equation of reactions and state the conditions.
(iv) Name one acid which when prepared using concentrated sulphuric acid ends in reacting with it.
Answer (c)
(i) Its strength as a strong acid and its low volatibility.
(ii) I. KNO3 and conc. H2SO4
II. NaCl and conc. H2SO4
(iii) I.
KNO3(s) + H2SO4(l) Heat
KHSO4(s) + HNO3(l)
Heat
II. NaCl(s) + H2SO4(l)
NaHSO4(s) + HCl(g)
(iv) Methanoic acid.
HCOOH(l) + H2SO4(l)
CO(g) + H2O(l) + H2SO4(aq)
(d) Sulphuric acid shows its typical acid behaviour by;
I.
Reacting with a carbonate or hydrogencarbonate to form a salt, water and carbon dioxide.
II. Reacting with a metal to form a salt and hydrogen
III. Reacting with a base to form a salt and water only.
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(i) What is the test for;
I. Hydrogen gas
II. Carbon dioxide
(ii) Write equations to illustrate reactions in each of the cases I, II and III using appropriate examples.
(iii)
Which of the reactions in I, II and III is neutralisation? Explain
(iv)
Why is it not possible to prepare salt.
I.
PbSO4 by reacting carbonate and acid?
II. PbCl2 by reacting carbonate and acid?
III. CuSO4 by reacting metal with acid?
IV. CaSO4 by reacting carbonate with acid?. Explain your answer in each
case.
Answer (d)
(i)
I. Burning splint produces a ‘pop’ sound with hydrogen.
II. Carbon dioxide forms a white precipitate with lime water.
(ii)
I.
H2SO4(aq) + Fe(s)
II. H2SO4(aq) + ZnCO3
III. H2SO4(aq) + CuO(s)
FeSO4(aq) + H2(g)
ZnSO4(aq) + CO2(g) + H2O(l)
CuSO4(aq) + H2O(l)
(iii)
Reaction III is neutralisation. It is a reaction with a metal oxide or metal hydroxide to form a salt and water only.
(iv)
I. Little reaction occurs between sulphuric acid and lead carbonate because insoluble PbSO4 formed prevents reaction.
II. Little reaction occurs as insoluble PbCl2 formed prevents reaction from occuring.
III. No reaction occurs between copper and sulphuric acid because copper is below hydrogen in reactivity.
IV. Little reaction occurs between calcium carbonate and sulphuric acid because the slightly soluble CaSO4 formed prevents
reaction from occuring.
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(e) In an attempt to identify anions in salts X, Y and Z the following tests and observations were done and recorded as below:
Test
Observation
Inference
The salts X, Y and Z
were each dissolved
and tested.
(i) X - white precipitate
(a) To each solution
which effervesces on
was added a few
adding acid and gas forms
drops of Ba(NO3)2
white precipitate with lime
solution followed by
water.
a few drops of dil.
HNO3
I
(ii) Y - white precipitate
which effervesces with a
pungent smelling gas on
II
addition of acid.
(iii) Z - white precipitate not
affected by acid.
(b) To precipitate with
III
White precipitate remains
Z above was added
IV
dil. Nitric acid
V
(c) To a solution of y
Orange solution turned green
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was added a few
drops of dichromate
solution.
Answer (e)
Inferences;
I.
X is likely to be a carbonate
II.
Y is likely to be a sulphite
III.
Z is confirmed to be a sulphate
IV.
Z is confirmed to have SO32-
V.
Y is confirmed to have reducing ion, SO32-
(f) State three uses of;
(i)
Sulphuric acid
(ii)
Sulphates
Answer (f)
(i) Manufacture of fertilisers e.g. ammonium sulphate, (NH4)2SO4, calcium hydrogenphosphate, Ca(H2PO4)2
- Treatment of metals to remove rust before processes like tinning or galvanising.
- Sulphuric acid is also used in accumulators for automobiles
(ii)
Barium sulphate, BaSO4 is used as a pigment for paints.
-
CaSO4 is used as a fertiliser mixed with calcium hydrogensulphate to produce the super phosphate fertiliser.
-
“Plaster of Paris”, (CaSO4)2.H2O is used for setting broken bones.
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PRACTICE QUESTIONS
1(a) The diagrams below represent the changes that take place when solid sulphur is heated to a temperature of about 4500C. Use
diagrams to answer the questions that follow.
1150C
A
1600C
B
C
1440C
D
(i) What happens to the sulphur at 1150C.
(ii) Write the formula of sulphur molecule in both solid and liquid states and describe its shape.
(iii) What happens to the sulphur at 1600C?
(iv) Write the formula of sulphur molecules in gas state.
(v) Describe the state of sulphur at stage C stating the molecular structure
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b) When hot runny liquid sulphur is poured into cold water it forms an allotrope of sulphur
(i) Name this allotrope
(ii) What changes will occur on this soft pliable substance soon afterwards. Name the allotrope formed.
(iii) How is this substance different from other allotropes of sulphur
(c) Study the diagram below and answer the questions that follow.
beaker
Boiling tube
Boiling water
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Solution of sulphur
In methylbenzene
heat
Tripod stand
(i)
Hot methylbenzene was first poured onto sulphur on a filter paper in funnel. What was the purpose of this?
(ii)
What will be seen in the boiling tube above 960C?
Q2. Some sulphur in powdered form was put on the end of a deflagrating
spoon and heated in a bunsen flame until it started
burning.
(a)
What will be observed?
(b)
Write equation of reaction.
(c)
The spoon was then lowered into a gas jar of oxygen. After burning stopped some water was added.
(d)
(i)
What change was seen on flame
(ii)
Name solution that was formed and write equation for reaction that occurred.
A few drops of litmus solution was added to the solution in (c) above.
What was the colour of solution.
(e)
Sodium hydroxide solution was then added to the solution drop by drop until the litmus changed colour.
(i) What was the new colour?
(ii) What reaction took place between sodium hydroxide and the solution?
(iii) Write equation of reaction with the sodium hydroxide.
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Q3. (a) Iron powder was added to sulphur powder and mixed thoroughly.
(i)
Suggest one way of separating iron out to leave sulphur and another way of getting sulphur out to leave iron.
(ii)
The mixture was heated and the iron glowed red and a black solid thought to have the formula FeS was formed.
I.
What change took place? Explain
II.
When FeS is burned in air, write equation of reaction.
(b) To the black solid, FeS was added dilute hydrochloric acid and effervescence of a gas with rotten egg smell of formula, H yS was
formed.
(i)
What is the most likely value of y?
(ii)
Write equation of reaction between iron sulphide and hydrochloric acid.
(c) The gas in (b) above was subjected to the following reactions
I.
It was burnt in oxygen
II.
It was shaken with potassium dichromate solution.
(i)
What was seen in each case?
(ii)
Write equation of reaction when hydrogen sulphide, HyS burns in air?
Q4. The diagram below shows stages in the manufacture of sulphuric acid by contact process.
C
Gas X
Gases cleaned
Gases heated
Gases
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to 4500C, and
And dried
Pressurised to
Gas Y
passed
over
2-3 atmospheres catalyst
D
Gas dissolved
In conc. H2SO4
E
Product diluted
With water
(a) Name gases X and Y
(b) Where do the gases X and Y react?
(c) Write equations of reaction for gases X and Y
(d) Name product dissolved at D and write equation of reaction
(e) What happens to 98% of gases?
(f) What is done with 2% of gases?
(g) Name catalyst at C.
(h) Write equation of reaction at E.
(i) Write equation of reaction between
(a)
Dil. Sulphuric acid and potassium sulphite
(b)
Dilute sulphuric acid and potassium sulphite
(c)
Magnesium sulphate and barium chloride solution.
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(j) State three uses of sulphuric acid
Q5. (a) In the contact process for manufacture of sulphuric acid.
(i) Name the starting materials and write equations for the changes they undergo.
(ii) What are the conditions for manufacture of sulphuric acid.
(iii) Explain other reactions that lead to the formation of concentrated sulphuric acid. Use any relevant equations.
(b) State three chemical properties of sulphuric acid
(c) Sulphuric acid from manufacturers in 98% sulphuric acid and it has a density of 1.84gcm3.
(i) Calculate its molarity
(ii) What volume will be used to make 250cm3 of 2M sulphuric acid solution.
(H=1.0, S=32.0, O=16.0)
(d) Write ionic equations for reactions between dilute sulphuric acid and;
(i)
Magnesium
(ii) Sodium carbonate
(iii) Sodium hydroxide solution
(iv) Barium nitrate solution
(e) Element X can be represented as
32
X
16
I.
Draw atomic structure of X
II. Write its electronic configuration.
(ii)
Use the grids below to place the elements Carbon, Nitrogen and Sulphur in the periodic table (Atomic nos. of carbon Nitrogen
and Sulphur are 6, 7 and 16 respectively).
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(iii)
Makesi mangi
Give the groups and periods of the elements.
Group
Period
Carbon
Nitrogen
Sulphur
CHAPTER V
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1. a) Define the terms below giving an example
i)
Functional group
ii)
Isomerism
iii)
Hornologous series.
Answers.
a) i) A functional group is the portion of the molecule that determines
a compounds properties; usually the non-hydrocarbon part of the
molecule.
Examples are - X ( =Cl, Br, I ) for halogenoalkanes.
- OH(for alcohols)
- H (alkanes), C = C (alkanes) or – C = C – alkynes
This last group of examples unique.
ii) Isomerism is the existence of a molecule with the same chemical
formula but different structure.
Examples – C4H10 (butane) can exist as n – butane
CH3 CH2 CH2 CH3
Or
2 - methylpropane
CH3 CH(CH3) CH3
iii) Homologous series is a group of compounds with the same chemical properties and same general formula. Physical
properties change gradually.
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Example: alkanes, CnH2n+2
b) Compare a group in the periodic Table and a homologous series.
Answer:
Group
Homologous series.
1. Each element has
1. Each compound has
different formula
different formula but
same general formula
2. Elements in vertical
2. Compound with same
column of periodic
functional group.
table with same no. of
outermost electrons
3. Compounds have same
3. Elements have same
chemical properties.
chemical properties
c) Write equations for the combustion of
i)
Ethene, C2H6
ii) Propane, C3H8
Answer:
i) 2C2H6 + 5O2(g)
4CO2(g) + 6H2O(l)
ii) C3H8(g) + 5O2(g) 3CO2 + 4H2O(l)
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d) State the conditions necessary for he following reactions of alkanes to occur.
i) Halogenation
ii) Combustion.
Answer
i) Sunlight (u. v. Light)
ii) Heat
e) i) Write equation to show dichlorination of propane, C3H8.
ii)
State one disadvantage of this type of reaction for synthesis of
halogenoalkanes.
iii)
Halogenation of alkanes occurs by substitution.
Explain this statement.
Answer.
i) CH3 CH2 CH3 + 2Cl2
CH3C(Cl)2 CH3 + 2HCl
ii) A mixture of substitution products will be obtained.
iii) When halogenation occurs replacement of hydrogen by halogen on the hydrocarbon chain occurs. This type of reaction
where one atom replaces another is called substitution.
f) Give the systematic names of the following.
i)
CH3, CH, CH3
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CH3
ii) CH3 CH2 Cl
iii) CH3 CH2 CH - CH 2 CH3
CH3
CH2
CH3
iv) CHCl3
Answer: i) 2 – methylpropane
ii) Chloroethane
iii) 3 – ethylhexane
iv) trichloromethane.
g) Draw the structural formula of the following compounds.
i)
n – pentane
ii) 1, 1, 3 - trimethylbutane
iii) 2, 3 – dichloro – 5 – methylhexane.
Answer.
i) CH3 - CH2 - CH2 - CH2 - CH3
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ii)
CH3
Makesi mangi
H
CH3 - C - C - CH3
CH3
CH3
iii) CH3 – CH – CH - CH2 - CH - CH3
Cl
Cl
CH3
2. a) The set –up below was used to prepare and study gas X..
Three test tubes of gas X were produced.
i) What was observed when a lighted splint was held at the mouth of
one test – tube?
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ii) A large test-tube was held over the test tube with the lighted glint above it. A little lime water was added to this test tube
and shaken after stoppering. What was observed?
iii) The other test-tubes were shaken after adding potassium monganate VII and Bromine solution in each case. What was
observed.
iv) Name gas X.
v) Write an equation for the preparation of gas X.
vi) Write an equation for the complete combustion of gas X.
vii) Name a suitable salt of sodium which can be used to prepare ethane using the same method for preparing gas X. write
equation of reaction.
viii) What does the method of gas collection tell us about gas X.
Answer.
a) i) The gas burnt with a blue flame
ii) The lime water formed a white precipitate
iii) No change in potassium manganate VII. In bromine water there was a
gradual slow decolourisation because of presence of light.
iv) Gas X is methane
v) CH3COONa(s) + NaOH(aq)
vi) CH4(g) + 2O2(g)
CH4(g) + Na2CO3(ag
CO2(g) + 2H2O(l)
vii) Sodium propanoate.
CH3CH2COONa(s) + NaOH(aq)
C2H6(g) + Na2CO3(ag
viii) Gas X is insoluble in water.
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b) Complete the following equations by giving the major organic product.
i) CH3Cl + H2 NaOH
Pd / C
ii) CH3 CH2 Cl + LiAlH4
Zn / Cu2+
iii) CH3 CH2Br
H+
iv) 2CH3I + 2Na ethoxyethane
v) CH3 CH2MgBr H2O
vi) CH3COONA Electrolysis
vii) AlC3 + HCl
Answer. 2b)
i) CH4
ii) CH3 CH3
iii CH3 CH3
iv) CH3 CH3
v) CH3 CH3
vi) CH3 CH3
vii) CH4
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c) State three uses of alkanes.
Answer. 2c)
- Methane is used as fuel gas and manufacture of other compounds (ethyne, methanol).
- Ethane is used in manufacture of tetraethyllead
- Higher alkanes used as petrol, fuel oils, vaseline etc.
3. a) Define the following terms giving an example:
i) Geometrical isomerism
ii) Hydrocarbons
iii) Insaturation
iv) Addition reaction.
Answer: 3(a) (i)
Geometrical Isomerism is the existence of a compound with two different structures due to restricted rotation about a carbon-carbon
double bond eg C is – 2 – Butene,
CH3
H
C
C
CH3
H
CH3
H
(trans – 2 – Butene,)
C
C
H
CH3
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ii) Hydrocarbons are compounds made up of carbon and hydrogen
only e.g propene C3H6 or propane C3H8
iii) Unsaturation is a situation in which carbon in a compound is not
attached to four other atoms e.g ethene ( H2C = CH2)
Ethyne ( Hc = CH ) are unsaturated.
iv) An addition reaction is one in which two or more compounds
combine to give a single product.
b) Name the following alkenes.
i) CH2 = CH Br
ii)
H
H
C
C
CH3
CH =
CH2
iii) CH2 = CH CH2C (CH3)3
iv) (CH3)2 C = CHCH3
v)
CH3 CH2
H
C
C
H
CH2 CH3
vi) CH3 CH2
H
C
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C
vi)
CH3 CH2
H
H
CH2
C
C
H
CH3 CH2
Answer.
i) Bromoethene
ii) C is Penta - 1, 3 – diene or C is 1, 3 - pentadiene
iii) 4, 4 - dimethyl – I - pentene
iv) 2 – methyl – 2 - butene
v) trans – 3 - Hexene
vi) C is – 3 - Hexene
vii) Trans – 2 – pentene.
c) Write structural formulae for
i) 2 - Methylpropene
ii) 2 – Methylbuta – 1, 3 - - diene
iii) Cyclobutene
iv) 4, 4 – dimethylpent – I - ene
v) 3 – Methylcyclopentene.
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Answer.
i) 2 - Methylpropene
CH3 - C = CH2
CH3
ii) CH -
CH2
CH
CH2
iii) 2 - Methylbuta 1, 3 – diene H2C = C - C = CH2
CH3
4, 4 – dimethyl – 1 – pentene
H
CH3 - C (CH3)2 – CH2 – CH = CH2
HC = CH
CHCH3 ( 3 – methylcydopentene)
H2C
CH2
d) Give the structural formula of isomers of pentene. Name them.
Answer:
CH3 – CH2 - CH = CH - CH3 ; 2 pentene
CH3 – CH2 - CH2CH = CH2
CH3 – CH - CH = CH2
; 2 pentene
; 3 – Methyl-I- butene
CH3
CH5 – CH2 – C = CH2
; 2 – methyl – I – butene
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CH3
CH3 – CH = C - CH3
; 3 methyl- 2 – butene
CH3
H2C _________
CH2
CH2
CH2
cyclobutane
CH2
H2C _________ CH2
H2C
CH
Methylcydobutane
CH3
H2C _________ CH2
Ethylcycloproppane
CH
CH2
CH3
Each has same formula, C5H10
e) State whether these are true or false and explain your answer.
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i) Alkenes are isomeric with cycloalkanes
ii) There is free rotation about carbon – carbon double bond.
Answer e(i) True. Alkenes have the general formula,
Cn H2n and so do cycloalkanes but structure differ
e.g cydohexane, C6H12 - cyclic
and hexene, C6H12 - Straight chain.
ii) False. Rotation about a double bond tantamount to breaking at least a carbon bond. This is not easy and does not occur.
4. a) The apparatus below were used to prepare ethene and collect it in gas
jars. Study diagram and answer question that follow.
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i) Name the excess acid Y
ii) At what temperature does the gas ethene begin to form?
iii) Name another product likely to form in this reaction.
iv) The first two or three gas jars to be filled were discarded. State the
reason.
v) State four physical properties of ethene
vi) One of the last gas – jars to be filled was selected and set on fire with a
burning splint
I.
Describe the flame and explain your answer
II.
State why it is important not to choose one of the first jars
III.
What is observed when lime water is placed in the gas-jar after all the gas has burnt up.
IV.
State any other observation
V.
Write equation for the complete combustion of ethene, C2H4
VI.
A few cm3 of bromine water was placed in another gas – jar of ethene, cover replaced and then gas jar shaken. State and
explain the observation
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VII. To the last gas jar was placed a few cm3 of dilute potassium permanganate (potassium manganate VII) solution to which
a little dilute sulphuric acid has been added. State the observations and explain.
Answer.
i)
Acid Y is conc. Sulphuric acid
ii) About 160 – 170oc
iii) Diethyl ether, CH3CH2 – O - CH2CH3
iv) They would contain air mainly
v) Ethene is a colourless gas, has sweetish smell, insoluble in water and has approximately the same density as air.
vi) I). Ethene burns with a yellow (luminous) and sooty flame. Alkenes
have a high proportion carbon and the decomposed carbon glow
in the heat of the flame producing the luminous nature. Flame is
rooty because some carbon escape unburnt.
II) The first gas-jars likely to contain air and a mixture of air and ethene is explosive.
III) Lime water formed a white precipitate.
IV) The inside of the gas-jar was cloudy and misty.
V) C2H4(g) + 3O2(g)
2CO2(g) + 2H2O(l)
VI) The red bromine water decolourised. An addition reaction occurs to form a non-coloured compound; 2 – dibromoethane.
CH2 = CH2 + Br2
(g)
Br CH2 - CH2Br
(g)
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VII) The purple potassium permanganate decolourised. Alkenes are easily oxidised and potassium permanganate is reduced to
a very pale pink compound.
b) The reaction process in (a) above is known as dehydration and it is an example of elimination reaction.
i) Define the terms.
I)
Dehydration reaction
II)
Elimination reaction.
ii) Complete the following reactions dehydration by giving the main organic product.
I. CH3 CH3 OH
95% H2SO4
1700C
II. CH3 CH2OH Al2 O3
300 – 3500C
III. CH3 CH2 OH
H3 PO4
200 - 2500C
iii) What are common to the three dehydration reactions?
Answer: (i)
I. Dehydration reaction is one in which the elements hydrogen and
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oxygen are removed from a compound in some proportion as they
are in water.
II. Elimination reactions lead to removal of atoms from a molecule and
formation of a double (or triple) bonds on ring structures occurs.
ii) In (ii) I, II and III the product is ethene, CH2 = CH2
iii) - High temperature required in all the reactions
- The catalysts are acids – sulphuric acid, phosphoric acid and lewis acid
(Al2 O3)
- Reactions start from ethanol and produce ethene and water.
c) Study the flow chart below and answer the questions that follow.
Crude Oil
Process A
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Gases including
insaturated
hydrocarbons
Makesi mangi
Heavier
hydrocarbon
fractions
Volatile liquid
hydrocarbons
Process B
General usages
and demand
General usages
and demand
Many uses but more
specialised and therefore
of relatively lower
demand
i) Name processes A and B
ii) I. Why is process B important?
III.
Complete the following simple equation which illustration process B
C5H12 H2O(g)
900oC
iii) Name a natural source of gaseous alkenes.
iv) State the demand and general usages of
I.
Gases and
II. Volatile liquids form crude oil.
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Answers.
C) i) A is fractional distillation and B is cracking
ii) I. Process B is important to supplement the supply of gaseaous and
liquid hydrocarbons which are on a relatively much higher
demand.
II) C5 H12
H2Og
C2H4(g) + C3H8(g)
9000C
iii) Crude oil or petroleum.
iv) I. Gases are used as cooking and lighting fuels and manufacture of
plastics by addition polymerisation. The demand is high.
II. Volatile liquids are used as liquid fuels e.g. petrol and paraffin.
The demand is very high.
5. a) Complete the following reactions by giving the main product(s)
i) CH2 = CH2 + H2
N
1400C
ii) CH2 = CH2 + Cl2
Ccl4
iii) CH2 = CH2 + O2
heat
iv) CH2 = CH2 + Hcl
(Markornikor product)
vi) CH3 - CH = CH2 + HCl
Peroxide
(Anti-markornikor product)
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Answers. 5a)
i) CH3 CH3
ii) Cl CH2 - CH2Cl
iii) CO2(g) and H2O(l)
iv) H3C - CH2Cl
v) CH3CHCl – CH3
vi) CH3 CH2 - CH2Cl
b) At high pressures of about 1000 atmospheres and a temperature of 2000C ethene molecules combine to form a high relative
molecular mass solid in a process known as polymerisation. Traces of oxygen act as catalyst. Ethene is known as monomer.
i) Define the following terms.
I.
Monomer
II.
Polymer
III.
Polymerisation
ii) Write equation for the polymerisation of ethene giving the conditions.
Answers.
i) I. Monomers are the small molecules that join up to build large molecular
compounds called polymers.
II. A polymer is a compound made up by joining many small molecules to
produce large molecules.
III. The process of joining up of small molecules to form a polymer is
known as polymerisation.
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ii) 2n CH2 = CH2
Makesi mangi
1500 atm - CH2 - CH2 - CH2 - CH2)n
200oC, O2
c) State five uses of alkenes.
Answers.
- Ethene is used for the manufacture of polythene and chloroethene – for manufacture of polychloroethene (polyvinylchloride)
- Propene is used to make polypropene for various plastics.
- Ethene is used to make refrigeration and ethanol.
- Butadiene is used to manufacture artificial rubber.
d) Give three uses for each of the following polymers.
i) Polyethene
ii) Polypropene
iii) Polychloroethene.
Answers:
i) Polyethene is used for making plastic bags, packaging and plastic films
ii) Polypropene is used to make ropes, buckets, food containers and
washing up bowls.
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iii) Polychloroethene is used to make raincoats, insulation for electric
cables, water pipes, suitcase coverings.
6. a) Below is an arrangement used to prepare ethyne.
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i) The rubber teat is used to squeeze out water which reacts with calcium
carbide (CaC2) producing the gas and calcium hydroxide. Write
equation of reaction.
ii)
Apart from air, impurities of H2S, PH3 or A5H3 are normally found in
gas. These give it a foul smell. How are these gases removed?
iii) State two physical properties of the gas you may deduce from the observations.
iv) Four test tubes of the gas were collected and gas tested as follows:
I. Put a lighted splint at the mouth of the last test tube filled with gas. Write observations and explain.
II. A few drops of bromine water were put in another test tube and the stopper replaced and shaken. State observations and
explain.
III.
A few drops of dilute acidified manganate VII were added to another test tube, stopper replaced and shaken. State
observations and explain.
IV.
To the last remaining test tube add a few drops of ammoniacal copper (I) chloride solution and replace stopper and
shake mixture. State observations and explain.
Answers: a) i)
i) 2H2O(l) + CaC2 (s)
Ca (OH)2(s) + C2 H2(g)
ii) There gases will dissolve in water especially when mixed together while ethyne is insoluble.
iii)Gas is colourless and insoluble in water.
iv) I. Gas burns with a yellow and very smoky flame. This is because ethyne has
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a very high proportion of carbon to hydrogen. This carbon is not burnt
fully producing soot.
II. The bromine water which is red brown becomes decolourised. Alkynes
are unsaturated and take part in addition reactions with halogens. The
colour takes a few minutes to disappear.
The purple solution decolourised to pale pink. Purple manganate VII ions reduced to Mn2+ ions in the presence of
III.
alkynes and Mn2+ ions are pale pink.
IV. A yellow precipitate forms. A terminal alkyne forms salts with metals. The hydrogen of inch alkyne is polar and gets
replaced by ions of metals like Cu+ and Ag+ to form salt precipitate.
b) i) Name the following compounds.
I.
(CH3)3 C CH2 C = C CH3
II.
CH3 CH Cl C = CH
ii) I. Give the formula of the first four alkynes.
II. Write the general formula for alkynes.
III. Give the structural formula of isomers of pentyne and name them.
iii) Give the structural formula for:I.
4 – Chlorobut – I – yne
II.
3 – Methylpent – 3 – en – 1 - yne
III.
5, 5 – dimethylhex – 2 – yne
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hexa – 1, 5 – dien – 3 – yne.
IV.
Answer. B (i) I.
II.
5, 5 – Dimethylhex – 2 – yne
3 – Chloro-I-butyne.
ii) I. C2H2 (ethyne), C3H4 (propyne)
C4 H6 (butyne), C5H8 (pentyne)
II. General formula is CnH8 (pentyne).
III. CH3 CH2 CH2 C = CH
1 – pentyne
CH3 C = C CH2 CH3
2 – pentyne
CH3 CH C = CH
3 – methyl-I- pentyne.
CH3
iii) I. H2C - CH2 – C = CH
Cl
II. CH3 – C = C - C = CH
H
CH3
CH3
III. CH3 - C - CH2 - C = C - CH
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CH3
IV. H2C = C - C = C - C = CH2
H
H
c) i) Define the terms.
I.
Unsaturation
II.
Saturation
ii) Illustrate your answer in c(i) above using reactions of
I.
Ethyne
II.
Ethene
III.
Ethane
Answer. c(i)
I.
Unsaturation is a state in which a compound has one or more double or
triple covalent bonds.
II.
Saturation is a state in which each of the carbon atoms in a compound
has for single covalent bonds.
ii) I. Ethyne is unsaturated because it has a triple bond. Unsaturated compounds undergo addition reaction:Br
H – C = C – H + 2Br2
Br
H
Br
C
C H
Br
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II. Ethene has a double bond. It will also undergo addition reaction.
H2C = CH2 + Br2
Br
H
H
C
H
Br
C H
III. Ethene is saturated and will therefore undergo substitution reaction
in the presence of heat or sunlight.
Br
H
H
C
H
H
CH3 - CH3 + Br2
C H
+ HBr
d) State two ways of distinguishing an alkene from an alkyne.
Answer.
- Alkynes are not oxidised by chronic acid (yellow) therefore no observable colour change on adding a few drops of chronic acid
and shaking. alkenes will change the colour form yellow to green.
- Oconolysis of alkynes yields carboxylic acids and hence product produces a gas with sodium carbonate or hydrogen carbonate.
Alkenes yield alkanols and alkanories with no effect on sodium carbonate or hydrogencarbonate.
- Alkynes decolourise bromine water only slowly taking a few minutes while with alkenes the reaction is instant.
e) State four uses of alkynes.
Answer;
- Manufacture of chloroethene for PVC
- Manufacture of trichloroethene, a solvent
- Ethyne is used in welding when mixed with oxygen
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- Manufacture of But-z-en-I-yne for synthetic rubber.
f) Study the reaction scheme below and answer the questions that follow.
Zn
E
D
A
KOH(alc)
NaNH2
2Br2
HC = CH
CH2
CH2
Br
Br
Na(s) or NaNH2
B
RX
C
Y
CH2 = CH2
i) Give reagent Y
ii) Give products A, B, C, D and E
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Answer: f) i) Bromine (Br2)
ii) A = CH = CH2
Br
B = HC = C Na
C = HC = C – R
D =
Br
H C
Br
Br
C -H
Br
E = H - C = C - H.
g) A sample of gas X occupying 120cm3 of formula C, Hy was burned in 360cm3 of oxygen (an excess). Residual gases occupied
300cm3 but after passing through sodium hydroxide solution only 60cm3 of gas was left. All volumes were measure at room
temperature and pressure conditions.
i) Write a general balanced equation for burning process for compound
X using x and y.
ii) What was the purpose of sodium hydroxide solution.
iii) Determine the values of x and y and hence the formula of compound X.
Answer. g ( i)
i) Cx Hy(g) + ( x + y) O2(g)
xCO3(g) + y2 H2O(l)
4
ii) Sodium hydroxide solution is used to absorb acidic gas – carbon dioxide.
iii) Vol. Of oxygen reacted = (360 – 60) = 300cm3
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= ( 300 – 60) = 240cm3
Vol. Of CO2 formed
Therefore vol. Ratio CxHy
: O2
: CO2
= 120
: 300
: 240
= 1
:
: 2
Therefore x
=
hence ( x + y )
2
2.5
(see equation)
= 2.5
4
Therefore 2 + y = 2.5
4
hence
y
= ( 2.5 - 2 ) x 4
= 2
therefore formula is C2 H2
PRACTICE QUESTIONS.
1. a) i) Propane, C3 H8 is an example of a hydrocarbon. Define the term
hydrocarbon.
ii) Write a general formula for alkanes and the formula of the sixth alkane.
iii) Name the main source of alkanes and the method of separation of the different molecules.
iv) Give the five structural isomers of hexane and name them.
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b) i) Name the type of bond and structure in alkenes
ii) Arrange the following in order of increasing boiling point:N – pentane, 2, 2 – dimethylpropane, n-butane and ethane.
iii) State and explain the trends in the following properties of alkanes.
I.
Melting point
II.
Boiling point
III.
Density.
iv) State and explain whether alkanes dissolve in :I.
Water
II.
Organic solvents like petroleum ether.
c) i) Alkanes are saturated compounds. Define the term saturated.
ii) Write balanced equations for the complete combustion of:
I.
butane, C4H10
II.
pentane, C5H12
iii) In a car engine, combustion is usually not complete:I.
State possible compounds in exhaust gases in a car
II.
What is the colour of soot and its chemical name.
d) In halogenation of alkanes we may have monosubstitution, disubstitution and other successive substitutions until all the
hydrogens are replaced.
i) Write equation to show monosubstitution of ethane (C2 H6) in
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presence of ultra-violet light or heat.
ii) Use your product in (i) above to show monosubstitution.
iii) Name the final product in halogenation of ethane.
e) Name the following compounds.
i)
CH3 CH2 CH2 CH2 CH3
ii) C (C2H5)4
iii) CH3 CH2 CH2 C (CH3)3
iv) CH (CH3)2 CH (CH3 ) CH3
v) CH3 at (C2H5) CH2CH2Cl
vi) (CH3)2 C - CH2
Br
Br
vii) CH3 CH2 CHBr2
viii)
CH3 CH2 CH - CH2CH2CH2Br
Br
f) Give the structural formula of:
i)
2 – Chloro – 4 - methylhexane
ii) 3 – ethyl – 3 – methylpentane
iii) 1, 2 – dimethylcyclopentane
iv) 1, 1, 1 – trichloroethane
2. a) State three physical properties of alkenes. Explain your answers.
b) Distinguish between the following terms.
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i) Saturated and unsaturated hydrocarbons
ii) Hydrolysis and hydration
iii) Addition and substitution reactions
iv) Geometrical isomers and structural isomers
c) Given two unlabelled test-tubes, one containing ethane and the other containing ethane. Describe two tests you may carry out
to identify the gas in each test-tube.
d) i) Draw the geometrical isomers of 3 – hexene labelling each
appropriately.
ii) Draw at least two structural isomers of 3 – hexene and name them.
3. a) Below is an arrangement of apparatus used to prepare a gas from
medicinal paraffin.
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Three test tubes of gas were collected.
i)
A lighted splint was used to burn gas Y in a test tube. State observations
ii) Lime water was added to the test-tube in a (I) above and then cooked and shaken. What was observed?
iii) Bromine water was added to another test-tube of gas Y. What was observed?
iv) To the third test tube was added to potassium manganate (VII) solution. What was observed?
v)
Use the observations in a(i) – (iv) to identify whether gas Y is an alkane or an alkene.
vi)
What suggests that the product has smaller molecules than the paraffin? Explain.
b) In a similar experimental arrangement using porcelain in place of pumice and ethanol in place of paraffin. Three large test tubes
of gas were collected.
i)
What is the use of porcelain?
ii) A lighted splint was used to ignite gas in a test tube. What was observed.
iii) To the test tube above was added lime water after the gas was burnt and then corked and shaken. What was observed?
iv)
To the last test tube was added potassium manganate (VII) solution. What was observed?
v)
Write equation for reaction that occurred on ethanol (CH3CH2OH)
c) What term is used to describe reactions in (a) and (b) above.
4. a) Define each of the following terms.
i)
homologous series
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ii)
Makesi mangi
polymer
iii) structural isomerism.
b) Draw all isomers of 2 – butene – both geometrical and structural. Name each of them.
c) Name each of the following compounds by their IUPAC names.
i) Br
Br
C
C
H
H
iii)
H2
H2C
H2C
ii) Br
H
C
C
H
Br
iv)
CH
CH
H2C
CH
CH
H2C
CHCH3
CH2
d) Give structural formula for each of the following.
i)
2 – methyl-2-butene
ii) 1, 4 – pentadiene
iii) 2, 4, 4 – trimethyl-I-pentene.
iv) Trans – e – butene.
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e) Give the products in the following reactions.
i) CH2 = CH2 + Br2
CCl4
ii) CH2 = CH CH3 + HBr (markornikor product)
iii)
CH2
H2C
CH + HCl
H2C
(markornikor product)
C CH3
CH2
iv)
H2C
CH2
CH + H2
HC
(Excess) Product
CH
CH
f) The following are compounds derived from ethene.
H2C
= C H CL
Chloroethene
H2C
= CH
Phenylethene
C6H5
F2 C = C F2 Tetrafluoroethene
They all form addition polymers like ethene.
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i) Complete the following addition polymerisation reactions.
I.
F
F
F
F
C
C +
C =
C
F
F
F
F
II. 2n H2C = C F
C6 H5
ii)
Name the polymers derived from each of the three monomers by adding the prefix “poly –
iii)
The polymer from chloroethene is commonly known as polyrinylchloride. Give at least three uses for it.
iv)
Define the term addition polymerisation.
5. Give the name of each compound shown below as a formula.
i)
H
H
C
ii)
H
H
H
H
C
iii)
C
H
H
H
H H
H
C
C
C
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H
iv)
H
H
H
C
v)
C
H
H
H
H
H H
H
H
H
C
C
C
Cl
H
H
H H
H
a) Which formula contains a double bond
b) Three of the six substances belong to the same family or homologous series. Name them.
c) Substance ( iv) was shaken with bromine water. State and explain the observations. In your explanation use the terms
“addition reaction” and “unsaturated compound”.
d) Compound (ii) reacts as follows presence of sunlight.
H
H
H
C
H
H
C
H
H
H
+ Cl2 Sunlight
H
H
C
C
C+ HCl
H
Use this equation to explain the terms “substitution reaction”.
6. a) Use the reaction below to answer questions that follow.
CH4(g) + 2O2(g)
CO2(g) + 2H2O, H = -888kJmol-
i) Name reactants
ii) Name products
iii) Is the reaction exothermic or endothermic? Explain
iv) Is the molecule CH4 Saturated Or Unsaturated? Explain
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v) State the meaning of H = -888kJ mol-1?
b) Write structural formula for three isomers of pentane, C5 H12. Name them.
c) The terms cracking, fractional distillation, hydrogenation hydrolysis and polymerisation are used to describe various processes.
Classify the processes into those that :
i)
Involve breaking down large molecules into several small molecules
ii) involve building up of larger molecules from a few or many small molecules
iii) I do not involve any chemical reaction at all.
d) Draw structural formulae for
i)
1, 1, 1 - trichloropropane
ii) butan 1, 3 – diane
iii) 2 – methylpropane.
7. a) Propane, C3 H8 is an example of a hydrocarbon
i)
Define the term hydrocarbon
ii)
Write the general formula for alkanes
iii) Give name and formula of sixth alkane.
iv) Name the main source of alkanes and the method of separation of the various alkanes.
b) Below is a list of formulae for some carbon compounds. The rectangles represent carbon chains, which are not necessarily the
same.
i)
C = CH2
ii)
CH3
H
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iii)
NH2
iv)
OH
I.
Which two are hydrocarbons?
II.
Which two would be decomposed by heat to an alkene
III.
Which one would give out ammonia when heated with soda lime.
IV.
Which compound would react by addition polymerisation to form a plastic.
c) The terms thermoplastic (thermosoftening) plastic and thermsetting polymer are used to describe heat effect on plastics.
Distinguish these two terms.
8. a) Below is a table of name, structural formula and boiling points of some
alkynes. Use the table to answer questions that follow.
Name
Structural Formula
Boiling point
(0C)
Ethyne
H - C = C - H
-75
Propyne
CH3 – C = C - H
-23
But – I – yne
CH3 CH2 - C = C - H 9
Pent – I – yne
CH3 CH2 CH2 C =
40
CH
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ii)
Makesi mangi
State and explain the type of bond in alkynes.
iii) Compare the density of ethyne with that of air.
( C = 12,
H = 1, N = 14, 0 = 32)
b) i) Alkynes take part in addition reactions. State the meaning of the terms
“addition reactions”.
ii)
Following is a reaction scheme for ethyne. Study it and answer the questions that follow:-
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I.
Name products A, B, C and D, E
II.
Name the catalyst C giving the reagent and the right temperature condition
III. What is the difference between reactivity of hydrogen iodide and hydrogen bromide to ethyne.
IV. What is the role of FeCl3 in chlorination of ethyne?
V.
Both bromine and bromine water will be decolourised by ethyne. State the difference in reactivity and products.
VI. Write a balanced equation for the complete combustion of ethyne to produce B.
c) State two uses of alkynes
d) i) Draw structure of
I)
3, 3 – Dimethyl – I – butyne
II) 3 – hexyne
ii) Name I. (CH3)2 CH C = CH
III.
HC = C ( CH2)4 CH3
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208