21st Century Chemistry
Structured Question in Topic 3
Metals
Unit 9-12
1.
The table below shows information about four different metals — A, B, C and D.
Metal
Cost
Corrosion Conductivity Strength
resistance
A
Low
Low
Medium
Medium
Density
Low
B
High
High
Good
High
High
C
Low
Medium
Medium
High
Medium
D
Medium
Medium
Good
Medium
Medium
Which metal is the most suitable for making
medal;
B
Any two of the following:
High corrosion resistance
High mechanical strength
High density
(b) a gate; and
C
Any two of the following:
Cheap
High mechanical strength
Medium density
(c) the metal base of a hot pot?
D
Medium price
Good conductivity (heat)
State TWO reasons in each case for your choice.
(a)
[1]
[1 x 2]
[1]
[1 x 2]
[1]
[1]
[1]
(9 marks)
2.
Consider the following metals:
Aluminium, tin, copper, lead, zinc and titanium
For each of the tasks listed below, choose ONE metal which is best for accomplishing the
task. Explain your choice in each case.
(a) Making space shuttle
Titanium
[1]
High mechanical strength / light
[1]
(b) Protecting food can from rusting
Tin
[1]
As a surface protecting material to prevent oxygen and water from reacting
with iron.
[1]
(c) Making drinking can
Aluminium
[1]
High corrosion resistance / light
[1]
(6 marks)
3.
The figure shows a piece of electrical wire.
(a)
(b)
(c)
(d)
Which metal is the most suitable for making electrical wire?
Copper
[1]
Suggest two essential properties of that metal which make it suitable for making electrical
wire.
Good conductor of electricity
[1]
Ductile
[1]
Account for the properties mentioned in (b) in terms of metallic bond.
Copper is a good conductor of electricity because it has mobile electrons.
[1]
Copper is ductile because copper atom layers can slide over each other
without breaking its structure.
[1]
Explain with equation, if any, what will happen if this wire is put into
(i)
dilute hydrochloric acid; and
No observable change as copper has very low reactivity.
[1]
(ii) silver nitrate solution.
Solution turns blue/ black powder forms/ copper dissolves
[1]
[1]
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
or,Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
(8 marks)
4.
The table below lists some information about three metals A, B and C.
Metal
A
B
(a)
(b)
Atomic number
12
Action of heating in
air
Action with dilute
sulphuric acid
Burns with bright
light
Colourless gas
bubbles give off
20
C
----
Burns with brick-red
No observable
flame
change
No observable
Reacts vigorously
change
for a short period of
time and then stops
To which group in the periodic table do A and B belong?
Group 2/ Group II
[1]
(i)
Write an equation for the reaction between A and dilute sulphuric acid.
(An ionic equation will NOT be accepted for this question.)
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)/ A(s) + H2SO4(aq) →
ASO4(aq) + H2(g)
[1]
(ii) Draw the electronic diagram for the gaseous product formed in (i) above,
showing electrons in the outermost shells ONLY.
[1]
(c)
(d)
(e)
Explain why metal B reacts with sulphuric acid for a while and then stops?
Because insoluble calcium sulphate is formed
[1]
which prevents it from further reaction.
[1]
Suggest what metal C might be.
Silver/ gold/ platinum
[1]
Explain, in terms of electronic structure, why metal B is more reactive than metal A.
Metal B has one more electronic shell than A. The outermost shell electrons
of metal B are further away from the nucleus than those of metal A.
[1]
Therefore losing outermost shell electrons from metal B is easier than from
metal A.
[1]
(8 marks)
5.
Some reactions of four metals A, B, C and D are shown below:
Reaction
(i)
A(s) + HCl(aq) → no reaction
(ii) B(s) + HCl(aq) → steady reaction to give hydrogen
(iii) A(s) + O2(g) → slow reaction when heated
(iv) D(s) + O2 (g)
oxide is formed
(v) B(s) + H2O(g) → slow reaction with steam
(vi) C(s) + H2O(g) → vigorous reaction with steam
(a)
(b)
(c)
Arrange the four metals in descending order of reactivity. Explain briefly.
C>B>D>A
[1]
From reactions (v) and (vi), C is more reactive than B because C reacts
vigorously with steam while B reacts slowly with steam.
[1]
From reactions (i) and (ii), B is more reactive than A because B reacts with
hydrochloric acid but A does not.
[1]
From reactions (iii) and (iv), D is more reactive than A because oxide of D
is formed on heating while oxide of A does not.
[1]
Another metal E reacts with dilute hydrochloric acid to give hydrogen. E also shows the
reaction
B(s) + E2+(aq) → B2+(aq) + E(s)
Where would you place E in the reactivity series among A, B, C and D? Explain briefly.
E is more reactive than A because E reacts with dilute HCl to give
hydrogen.
[1]
E is less reactive than B because B can displace E from the solution of E.
[1]
Thus ,E should be placed between B and A in the reactivity series.
[1]
(i)
What does the sign "
" meant?
The reaction is reversible.
[1]
(ii) Suggest what D might be.
Mercury
[1]
(9 marks)
6.
(a)
(b)
For each of the following experiments, state ONE observable change and write a
chemical equation for the reaction involved.
(i)
Magnesium is put into dilute hydrochloric acid.
Magnesium dissolves/ bubbles are given out.
[1]
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
[1]
(ii) Sodium is heated under Bunsen flame.
Golden yellow flame is observed/ white solid is left.
[1]
4Na(s) + O2(g) → 2Na2O(s)
[1]
(iii) Lead(II) oxide is heated with carbon powder.
Yellow solid turns orange/ silvery solid is formed.
[1]
[1]
2PbO(s) + C(s) → 2Pb(s) + CO2(g)
(iv) Zinc is put into copper(II) sulphate solution.
Zinc dissolves/ brown solid deposits/ solution turns from blue to pale
blue.
[1]
Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
[1]
Explain why the there is no reaction in the following experiments.
(i)
Lead is put into dilute sulphuric acid.
Insoluble lead sulphate is formed
[1]
which prevent further reaction.
[1]
(ii) Calcium oxide is heated with carbon powder.
Calcium oxide is very stable
[1]
which cannot be reduced by carbon.
[1]
(iii) Copper is put into magnesium nitrate solution.
Copper is less reactive than magnesium,
[1]
copper cannot displace magnesium from its solution.
[1]
(14 marks)
7.
Sodium can be used as a drying agent to remove trace of water in organic solvents. Sodium is
first drawn into a wire and placed in a bottle of organic solvent.
(a)
(b)
(c)
(d)
Explain, with the help of equation, why sodium can be used as a drying agent.
Sodium reacts with water to give sodium hydroxide and hydrogen.
[1]
[1]
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Explain why sodium can be drawn into wire in terms of its structure.
Since the metal atoms in sodium are in layers structure,
[1]
the atom layers can slide over each other without breaking its structure.
[1]
When sodium was drawn into wire, it appears shiny at first but turns grey after exposed
to air. Explain this observation with equation.
When sodium is freshly drawn, new surface is exposed as shiny surface. But
it is rapidly oxidized by air to form oxide.
[1]
4Na(s) + O2(g) → 2Na2O(s)
[1]
Suggest a potential hazard of using this method in drying organic solvent.
Sodium may react explosively with other substances such as water, in the
process of preparation and disposal.
[1]
(7 marks)
8.
The reactivity of metals can be determined by the reaction between metal and acid.
Five different metals, calcium, tin, zinc, copper and nickel are added to dilute hydrochloric acid
in the following set-up. The gas produced is collected by displacement of water. The height of
gas collected is measured.
The results are shown in the following table.
(a)
(b)
(c)
(d)
(e)
(f)
Metal
calcium
tin
zinc
Height (in cm) of gas collected in 30
seconds
5.5
1
4.5
copper nickel
0
2.5
Write the chemical equation for the reaction between zinc and hydrochloric acid.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
[1]
Suggest a test for the gas evolved.
Hydrogen gas gives a 'pop' sound
[1]
with a burning splint.
[1]
Determine the reactivity series in descending order from the result provided.
Calcium > zinc > nickel > tin > copper
[1]
Explain why the reactivity of sodium cannot be determined by this method.
Sodium reacts with acid explosively.
[1]
Explain why it is not suitable to use sulphuric acid in determining the reactivity of metal in
this way.
Calcium will form insoluble sulphate with sulphuric acid
[1]
which prevent the metal from further reaction.
[1]
Two other metals were tested and their results were as follow:
Metal
Metal X
Metal Y
Height (in cm) of gas collected in 30 seconds
5
2
From the result obtained, rewrite the reactivity series and suggest what metals X and Y might
be.
Calcium > metal X > zinc > nickel > metal Y > tin > copper
[1]
Metal X: Magnesium/ aluminium
[1]
Metal Y: Iron
[1]
(10 marks)
9.
This following table shows the density of two Group I metals.
Metal
Density (g/cm3)
Li
Rb
0.5
1.53
3
(a)
Density of water = 1.0 g/cm
(i)
Suggest TWO observable differences when lithium(Li) and rubidium(Rb)
added into a water trough respectively.
Rubidium will sink in the water trough but lithium will float on water.
are
[1]
Rubidium reacts more vigorously than lithium.
[1]
Account for the differences.
Rubidium is denser than water (1.0 g/cm3) and lithium is less dense
than water.
[1]
Outermost shell electron in rubidium is further away from nucleus
than that in lithium.
This results in weaker attraction between nucleus and electrons. The
electrons in rubidium lose more readily.
[1]
(iii) Write the chemical equation for the reaction between rubidium and water.
2Rb(s) + 2H2O(l) → 2RbOH(aq) + H2(g)
[1]
Rubidium(Rb) is stored in paraffin oil in a bottle with suitable hazard warning labels.
(i)
Why should rubidium be stored in paraffin oil?
It prevents rubidium from reacting with moisture and oxygen in air.
[1]
(ii) Draw a hazard warning label that should attach on the bottle containing rubidium.
(ii)
(b)
Explosive
[1]
or,flammable
(7 marks)
10.
A metal, M reacts with very dilute nitric acid to form solution B and a gas C but has no reaction
with steam. When it is heated with Bunsen flame, it gives a solid D. The solid is orange when it
is hot and yellow when it is cold.
(a) Suggest what M, B, C and D might be.
M: lead
[1]
B: lead(II) nitrate
[1]
C: hydrogen gas
[1]
D: lead(II) oxide
[1]
(b) Suggest a test for gas C.
Gas C: It gives a 'pop' sound with a burning splint.
[1]
(c) Write balanced equation(s) for extraction of metal M from its sulphide compound.
2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)
(d)
[1]
[1]
2PbO(s) + C(s) → 2Pb(s) + CO2(g)
(i)
Describe what will be observed when a piece of metal M is added to silver nitrate
solution.
Lead dissolves.
[1]
Grey solid deposits.
[1]
(ii) Write an ionic equation for the reaction.
[1]
Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s)
(10 marks)
11.
The following table shows some information of metal reactivity series.
Metal
Extraction method
Ease of extraction
Difficult
K
Na
Ca
Method A
Mg
Al
Zn
Fe
Pb
Carbon reduction
Cu
Hg
Ag
(a)
(b)
(c)
(d)
(e)
(f)
Heating alone
Ease
Briefly describe the method A used in the extraction of reactive metals.
Electrolysis of molten ore
[1]
Metal will be obtained at cathode (negative electrode).
[1]
Write the equation for the carbon reduction used in the extraction of copper from
copper(II) oxide.
[1]
2CuO(s) + C(s) → 2Cu(s) + CO2(g)
(i)
Name another reducing agent that can be used instead of carbon to reduce
copper(II) oxide.
Hydrogen/ carbon monoxide
[1]
(ii) Write the equation for the reaction.
CuO(s) + H2(g) → Cu(s) + H2O(l)
[1]
or,CuO(s) + CO (g) → Cu(s) + CO2(g)
Write the equation for obtaining silver from its oxide by heating alone.
[1]
2Ag2O(s) → 4Ag(s) + O2(g)
Explain whether reactions occur in the following cases, state ONE observation, if any.
(i)
Put zinc into silver nitrate solution
Reaction occurs because zinc is more reactive than silver.
[1]
Zinc dissolves / grey solid deposits.
[1]
(ii) Put copper into magnesium nitrate solution
No reaction occurs because copper is less reactive than magnesium.
[1]
State the relationship between year of discovery and the ease of extraction for metals.
The easier the extraction method of a metal, the earlier the metal is
discovered.
[1]
(10 marks)
12.
The following metals arranges in the order of decreasing reactivity.
Metal a ; metal b ; metal c ; iron ; metal d ; metal e ; metal f
Based on the information given above, answer the following questions.
(a) Which metal is most likely to tarnish rapidly in air? Explain your answer.
Metal a
Because it is the most reactive metal and thus reacts rapidly with oxygen in
air to form an oxide layer on the metal surface.
(b) Which metal is most likely to be found free in nature? Explain your answer.
Metal f
Because it is the least reactive metal and thus does not form stable
compounds readily.
(c) Would you expect metal d react with cold water? Explain your answer.
No.
Iron has no reaction with cold water and metal d is less reactive than iron.
[1]
[1]
[1]
[1]
[1]
[1]
(d)
(e)
(f)
(g)
Suggest how metal a can be extracted from its ore.
Electrolysis of its molten ore
What would be formed when the oxide of metal d is heated with coke?
Metal d is formed.
Suggest an experiment to show that metal b is more reactive than iron.
Add metal b to a solution of iron (II) sulphate
Metal b can displace iron from iron (II) sulphate solution.
Can we store a solution of nitrate of metal c in a container made of metal d?
Explain your answer.
Yes.
Since metal d is less reactive than metal c, no displacement reaction occurs.
[1]
[1]
[1]
[1]
[1]
[1]
(12 marks)
13. A student performs several experiments to determine the order of reactivity of five metals (A, B,
C, D and E). The results are shown in the table below.
Metal
Experiment
A
B
C
D
E
Reacts
Reacts
No reaction
No reaction No reaction
Reaction with water
vigorously
readily
Reacts very
Reacts
Reacts
No reaction
No reaction
Reaction with steam
slowly
vigorously
readily
React
Reacts
Reaction with dilute
Reacts
No reaction
No reaction
slowly
explosively
hydrochloric acid
readily
Metal E
Metal B
Metal C
Heating metal oxide
No reaction
No reaction
obtained
obtained
obtained
with carbon
Displacement
E displaces metal B from a solution of nitrate of B
reaction
(a) Arrange the metals in order of reactivity, starting with the most reactive one. Explain your
answer briefly.
D, A, C, E, B
[1]
A and D are more reactive than B, C, and E because A and D react with
water while B, C and E do not.
[1]
D is more reactive than A as D reacts more vigorously with water than A.
[1]
C is more reactive than B and E because C reacts slowly with dilute
hydrochloric acid while B and E show no reaction with acid.
[1]
E is more reactive than metal B as E displaces B from a solution of nitrate
of B.
[1]
(b) (i)
Suggest what metal A might be.
Calcium
[1]
(ii) Write a chemical equation for the reaction between metal A and dilute
hydrochloric acid.
Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g)
[1]
(iii) State TWO observations when metal A is heated in air.
It burns with a brick-red flame.
[1]
White solid is formed.
[1]
(c) (i)
Suggest what metal D might be.
Sodium / potassium
[1]
(ii) Write a chemical equation for the reaction between metal D and water.
[1]
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
(d)
or, 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
(iii) Suggest how we can extract metal D from its ore.
Electrolysis of its molten ore
(i)
During reaction, C forms C2+ ion. Write a chemical equation for the
between C and steam.
C(s) + H2O(g) → CO(s) + H2(g)
(ii)
[1]
reaction
[1]
Draw electron diagrams of the products formed in (i) above, showing the
electrons in the outermost shells only.
[1]
(e)
Metal B forms an oxide with a chemical formula of B2O.
(i)
Suggest what metal B might be.
Silver
[1]
(ii) Would there be any reaction between zinc and a solution of nitrate of B? Write an
ionic equation for the reaction, if any.
Yes
[1]
+
2+
[1]
Zn(s) + 2Ag (aq) → Zn (aq) + 2Ag(s)
(18 marks)
14.
The following tables record the results of some reactions of four different metals and their
oxides.
Metal
W
Reactions
X
No
Adding dilute hydrochloric Very dangerous!
observable
Never attempt
acid
change
Adding copper(II) nitrate
solution
Oxide
Reactions
Bubbles given
off
Oxide of W
No
observable
change
Oxide of X
Y
Z
Bubbles
given off
Bubbles
given off
Metal
dissolves,
brown solid
deposits
Metal
dissolves,
brown solid
deposits
Oxide of Y
Oxide of Z
No observable Silvery solid is No observable No observable
change
formed
change
change
Silvery grey
No observable
Heating with
No observable
----solid is formed
change
carbon
change
Arrange the four metals in descending order of reactivity. Explain your answer briefly.
W>Z>Y>X
[1]
Oxide of X can be reduced easily by heating only, so it is the least reactive.
[1]
Metal Y can be obtained by carbon reduction with its oxide, but W and Z do
not. It means that W and Z are more reactive.
[1]
Metal W is the most reactive, since metal W reacts vigorously with dilute
hydrochloric acid that we can never attempt.
[1]
If Z burns in air with a bright light, suggest what metal Z might be and write an equation
for its reaction with dilute hydrochloric acid.
Magnesium.
[1]
[1]
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Draw the electronic diagrams of the products formed in (b).
Heating alone
(a)
(b)
(c)
MgCl2:
[1]
[1]
H2:
(d) Explain the observation, with the help of equation, when metal W is added into copper(II)
nitrate solution.
W is so reactive that it reacts with water instead of displacing copper(II)
ions.
[1]
2W(s) + 2H2O(l) → 2WOH(aq) + H2(g)
[1]
(e) Suggest and explain an observable change when metal Y is put into copper(II) nitrate
solution.
Blue colour of the solution fades
[1]
as the concentration of copper(II) ions decreases.
[1]
(12 marks)
15.
The results of a series of displacement reactions are given in the table below.
Lead(II)
Iron(II)
Salt solution
Silver nitrate
Nitrate of X
nitrate
nitrate
Metal
X
---reaction occurs reaction occurs reaction occurs
no observable
no observable
reaction occurs
----change
change
Iron
no observable
reaction occurs
---reaction occurs
change
Silver
no observable no observable no observable
---change
change
change
What is displacement reaction?
The reaction of more reactive metal displaces less reactive metal ions from
its salt solution.
[1]
List two observable changes when X is put into iron(II) sulphate solution?
Any two of the following:
[1 x 2]
Metal X dissolves.
Brown solid deposits.
Green solution fades.
Based on the information given above, arrange the metals in descending order of
reactivity. Briefly explain your answer.
X > iron > lead > silver
[1]
X is the most reactive, since it can displace all other three ions from their
solutions.
[1]
Iron is more reactive than lead and silver, it can displace lead(II) ions and
silver ions from their solutions.
.
[1]
Silver is the least reactive as it cannot displace any metal ions at all.
[1]
Chromium should be placed between X and iron in reactivity series. You are provided with
chromium(III) nitrate solution, describe an experiment to show it.
Put X and iron into two separate test tubes with chromium(III) nitrate
solution.
[1]
X can displace chromium from chromium(III) nitrate solution but iron metal
cannot.
[1]
It shows that chromium is in between iron and lead in the reactivity series.
Write an ionic equation for the reaction between iron and silver nitrate solution.
[1]
Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)
(10 marks)
Lead
(a)
(b)
(c)
(d)
(e)
16.
(a)
(b)
Calculate the number of moles of the following substances.
(i)
1.2 x 1025 sodium atoms
1.2 x 1025 ÷ 6.02 x 1023 = 19.9 mol
(ii) 22.8 g nitrogen dioxide molecules
22.8 ÷ (14.0 + 2 x 16.0) = 0.5 mol
(iii) 4.8 x 1023 formula units of copper(II) carbonate
4.8 x 1023 ÷ 6.02 x 1023 = 0.8 mol
(iv) 13.2 g of hydrated magnesium sulphate, MgSO4･7H2O
13.2 ÷ [24.0 + 32.0 + 4 x 16.0 + 7 x (2 x1.0+16.0)]= 0.054 mol
Calculate the masses of the following substances.
(i)
2 moles of sodium hydroxide, NaOH
2 x (23.0 + 1 + 16.0) = 80 g
(ii) 1.9 x 1024 ethane molecules, C2H6
(1.9 x 1024 ÷ 6.02 x 1023) x (12.0 x 2 + 1.0 x 6) = 94.7 g
(iii) 0.4 moles of hydrated sodium carbonate, Na2CO3•10H2O
0.4 x [2 x 23.0 + 12.0 + 3 x 16.0 + 10 x (2 x 1.0 + 16.0)] = 114.4 g
[1]
[1]
[1]
[1]
[1]
[1]
[1]
(iv)
22
3.90 x 10 argon atoms
(3.90 x 1022 ÷ 6.02 x 1023) x 40 = 2.59 g
[1]
(Relative atomic masses: H = 1.0; C = 12.0; N = 14.0; O = 16.0; Na = 23.0; Mg = 24.0; S =
32.0; Ar = 40.0;Avogadro constant = 6.02x1023)
(8 marks)
17.
(a)
In 67.5 g of hydrated aluminium sulphate Al2(SO4)3•6H2O, calculate
(i)
the number of moles of aluminium ions, Al3+;
Number of moles of Al2 (SO4)3•6H2O
= 67.5 ÷ [ 2 x 27.0 + 3 x (32.0 + 4 x 16.0) + 6 x (1.0 x 2 + 16.0)]
= 0.15 mol
Number of moles of aluminium ions = 2 x 0.15 = 0.3 mol
(ii) the number of moles of water molecules;
Number of moles of water molecules = 6 x 0.15 = 0.9 mol
(iii) the number of sulphur atoms;
Number of sulphur atoms = 0.15 x 3 x 6.02 x 1023 = 2.71 x 1023
[1]
[1]
[1]
[1]
(iv)
[1]
the number of ions.
Number of ions = 5 x 0.15 x 6.02 x 1023 = 4.52 x 1023
[1]
When a metal M is completely changed to its ion, 1.96 g of M give 3.60x1022 electrons.
The atomic mass of M is 65.4.
(i)
What is the number of moles of 1.96 g of M?
Number of moles of M = 1.96 ÷ 65.4 = 0.03 mol
[1]
(ii) What is the number of moles of electrons given out?
Number of moles of electrons = 3.60 x 1022 ÷ 6.02 x 1023 = 0.06 mol
[1]
(iii) What is the charge on the ions of M?
The charge on the ions of metal M = 0.06 ÷ 0.03 = +2
[1]
(Relative atomic masses: H = 1.0; O = 16.0; Al = 27.0; S = 32.0)
(9 marks)
(v)
(b)
the number of oxygen atoms; and
Number of oxygen atoms = 18 x 0.15 x 6.02 x 1023 = 1.63 x 1024
18.
(a) 2.7 g of a metal M combine with 2.4 g of oxygen to form an oxide with the
M2O3. What is the relative atomic mass of M?
Let the atomic mass of metal M be Mr.
2.7
[1]
number of moles of M : number of moles of O = 2 : 3 = Mr :
(b)
19.
formula
Mr = 27
[1]
or, Mass of M in compound : Mass of oxygen in compound
= 2.7 : 2.4 = 2Mr : 16.0 x 3
Mr = 27
A crystalline salt (MSO4•7H2O) is found to contain 51.2 % by mass of water of
crystallization. Calculate
(i)
the formula mass of the hydrated salt; and
(ii) the relative atomic mass of the metal M.
(Relative atomic masses: H = 1.0; O = 16.0; S = 32.0)
Formula mass of the hydrated salt = (7 x 18.0) ÷ 0.512
[1]
= 246.1
[1]
Relative atomic mass of metal M = 246.1 - (7 x 18.0 + 32.0 + 4 x 16.0)
[1]
= 24.1
[1]
(6 marks)
(a) 5.1 g of hydrated calcium chloride (CaCl2•nH2O) gives 5.0 g of water on strong heating.
Find the value of n.
Number of moles of CaCl2 : Number of moles of H2O = 1 : n
[1]
Number of moles of CaCl2 = 5.1 ÷ (40.0 + 2 X 35.5) = 0.046 mol
Number of moles of H2O = 5.0 ÷ 18.0 = 0.28 mol
[1]
1/n = 0.046/0.28
n = 6.08 = 6
[1]
(b) A hydrated salt of copper contains 63.9% by mass of the anhydrous salt. The anhydrous
salt has the following percentage composition by mass: copper 39.8%; sulphur 20.0%;
oxygen 40.2%. Find the empirical formula of the hydrated salt.
Suppose there are 100 g of anhydrous salt.
Copper
Mass of element
39.8 g
Number of moles of 39.8 ÷ 63.5
atom that combine = 0.627 mol
Simplest ratio of 0.627 ÷ 0.625
= 1.003
atoms
Sulphur
20.0 g
20.0 ÷ 32.0
= 0.625 mol
0.625 ÷ 0.625
=1
Oxygen
40.2 g
40.2 ÷ 16.0
= 2.513 mol
2.513 ÷ 0.625
= 4.02
So the empirical formula of the anhydrous salt is CuSO4.
Formula mass of hydrated salt = (63.5 + 32.0 + 4 x 16.0) ÷ 0.639 = 249.6
[1]
[1]
[1]
Let the formula of the hydrated salt be CuSO4•nH2O.
63.5 + 32 + 4 x 16 + 18n = 249.6
[1]
n=5
The empirical formula of the hydrated salt is CuSO4•5H2O.
[1]
(Relative atomic masses: O = 16.0; S = 32.0; Cl = 35.5; Ca = 40.0; Cu = 63.5)
(8 marks)
20.
The following diagram shows the apparatus used in an experiment to determine the formula of
black copper oxide.
Town gas (main constituents are carbon monoxide and hydrogen) is passed over the oxide
before heating. During heating, the colour of copper oxide changes from black to brown. After
some time, heating is stopped, town gas was allowed to pass over the copper until it was cold.
The results are as follow:
Mass of tube
= 25.20 g
Mass of tube and copper oxide
= 27.78 g
Mass of tube and copper formed
= 27.26 g
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(Relative atomic masses: O = 16.0; Cu = 63.5)
Why is it needed to pass the town gas over the oxide before and after the experiment?
Before experiment: To flush the air out because a town gas and air mixture may explode
on ignition.
[1]
After experiment: To prevent oxidation of the copper by air again.
[1]
What is the mass of copper oxide used?
27.78 - 25.20 = 2.58 g
[1]
What is the mass of copper left?
27.26 - 25.20 = 2.06 g
[1]
Write appropriate equations to account for the change.
[1]
CuO(s) + CO(g) → Cu(s) + CO2(g)
[1]
CuO(s) + H2(g) → Cu(s) + H2O(l)
Calculate the number of moles of copper formed.
The number of moles of copper formed = 2.06 ÷ 63.5 = 0.0324 mol
[1]
Calculate the number of moles of oxygen that combined with copper in the copper oxide.
The number of moles of oxygen = (2.58 - 2.06) ÷ 16 = 0.0325 mol
[1]
From the results of (e) and (f), determine the empirical formula of copper oxide.
Mole ratio of copper and oxygen = 0.0324:0.0325 = 1:1
So the empirical formula of copper oxide is CuO
.
[1]
List TWO potential hazards of this experiment.
Town gas is toxic (carbon monoxide) and explosive (hydrogen).
[1, 1]
(11 marks)
21.
A piece of copper weighed 26.78 g is immersed in nitrate solution of metal X in a beaker. A
silvery deposit of X is formed on the copper surface and the solution gradually turns pale blue.
The beaker was then allowed to stand. X is then washed with distilled water and dried. X
weighs 1.94 g and the remaining copper weighs 26.21 g.
(Relative atomic masses: Cu = 63.5; X = 108)
(a) Explain why the solution gradually turns blue.
Since copper reacts and copper(II) ions is formed which is blue.
[1]
(b) Calculate the mass of copper reacted with nitrate of X.
26.78 - 26.21 = 0.57 g
[1]
(c) (i)
Explain why the solution is allowed to stand.
It allows the solid formed to settle and deposit on the bottom of beaker.
[1]
(ii) Explain why the solid X should be washed and dried.
Washing helps to remove any water soluble impurities.
[1]
Drying allows the water in solid X to evaporate off, so more accurate
result can be obtained.
[1]
(d) What is the number of moles of copper reacted?
0.57 ÷ 63.5 = 8.98x10-3 mol
[1]
(e) What is the number of moles of X formed?
1.94 ÷ 108 = 0.018 mol
[1]
(f) What is the number of moles of X reacted when 3 moles of copper(II) ions are formed.
(g)
Mole ratio of X and Cu = 0.018 : 8.98x10-3 = 2 : 1
Number of moles of X reacted = 2 x 3 = 6 mol
Write an ionic equation for the above reaction.
Cu(s) + 2X+(aq) → Cu2+(aq) + 2X(s)
[1]
[1]
[1]
(10 marks)
22.
A student tried to extract zinc from zinc oxide. He placed 1 g of zinc oxide and 10 g of carbon
powder in a crucible and heated with a Bunsen flame.
(Relative atomic masses: C = 12.0; O = 16.0; Zn = 65.4)
(a) Draw the experimental set-up for the extraction of zinc oxide.
(b)
(c)
(d)
[2]
Write the equation of the reaction involved.
2ZnO(s) + C(s) → 2Zn(s) + CO2(g)
[1]
(ii) Suggest one observation for the reaction.
The solid changes from white to yellow/ grey solid is formed.
[1]
(i)
Determine which reagent is in excess.
Number of moles of ZnO = 1/ (65.4 + 16.0) = 0.0123 mol
[1]
Number of moles of C = 10/12.0 = 0.833 mol
[1]
According to the equation, 2 moles of ZnO react with 1 mole of C.
Hence, 0.0123 moles of ZnO react with 0.00615 moles of C.
Therefore, carbon is in excess.
[1]
(ii) Calculate the maximum mass of zinc that can be obtained.
Maximum mass of zinc obtained = 0.0123 x 65.4 = 0.8044 g
[1]
In fact, the students only obtained 0.6 g of zinc.
(i)
Calculate the yield of the reaction.
Yield of reaction = (0.6 ÷ 0.8044) x100%
[1]
= 75.6%
[1]
(ii) Suggest one reason why the student cannot get the maximum mass of
zinc.
Temperature is not high enough/ time allowed for the reaction is not
long enough.
[1]
(11 marks)
(i)
23.
Explain each of the following:
(a) Aluminium does not react with steam unless when aluminium is washed with mercury(II)
chloride.
An aluminium oxide layer is attached on the surface of aluminium to
prevent its reaction.
[1]
Mercury(II) chloride can remove this oxide layer and allows aluminium
underneath to react.
[1]
(b) Silver nitrate solution cannot be kept in an iron bucket.
There is displacement reaction between silver nitrate solution and iron.
[1]
It will cause iron to dissolve.
[1]
(c) Aluminium, rather than copper, is used in making electrical cables.
[1, 1]
Aluminium is lighter and stronger than copper.
(d) Zinc can be extracted from its molten ore by electrolysis. Yet in practice, it is usually
extracted by heating its oxide with coke.
Electrolysis is an expensive method to extract metal.
[1]
(e) Galvanized iron is not used in making food can.
Zinc ion is toxic.
[1]
(8 marks)
24.
(a)
The above figure shows a factory which extracts zinc from its ore, zinc blende (ZnS). The flow
diagram below shows how zinc is extracted.
ZnS
Stage I
compound X and acidic
gas Y are formed
Stage II
Zn
In stage I, the ore is heated strongly in air. A yellow compound X is formed which becomes
white when cold. Besides, an acidic gas Y is evolved which is emitted to the surroundings from
the chimney.
(i)
(1) Write a chemical equation for stage I.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
[1]
(2) Give the names of compound X and gas Y.
X:
Zinc(II) oxide
[1]
Y:
Sulphur dioxide
[1]
(ii) In stage II, X is heated strongly with a black powder Z.
(1) What is Z?
Carbon
[1]
(2) Write a chemical equation for stage II.
2ZnO(s) + C(s) → 2Zn(s) + CO2(g)
[1]
(iii) State one use of zinc.
Making galvanized iron/ outercase of dry cell
[1]
(b) Statues A and B are made of iron mainly. It is found that statue A rusts more quickly than
B.
(i)
Give a reason for this.
Wind brings acidic sulphur dioxide gas to statue A but not B.
[1]
Acidic gas speeds up corrosion of iron.
[1]
(ii) Suggest one method to slow down corrosion.
Metal plating
[1]
(9 marks)
25.
Explain briefly how the protection against rusting can be achieved in the following cases.
(a) Underground iron gas pipe is jointed to magnesium.
This is sacrificial protection. Magnesium corrodes instead of iron
[1]
because magnesium is higher than iron in the reactivity series.
[1]
(b) Cloth hangers are coated with plastic.
Plastic acts as a protective layer
[1]
to exclude air and moisture and hence prevent rusting.
[1]
(c) Bus stop sign is painted.
Paint gives a protective layer to iron
[1]
so that iron can be kept away from direct contact with air and moisture.
[1]
(d) Bicycle chain is greased.
Grease protects the moving iron parts
[1]
by preventing air or water in contact with the iron.
[1]
(e) Car body is connected to the negative terminal of battery.
The battery of negative terminal releases electrons
[1]
which prevent iron from losing electrons.
[1]
(10 marks)
26.
Three test tubes are set up in the diagram shown below:
The tubes are left for three days.
(a) Which tube(s) will rusting occur?
C
[1]
(b) What is the function of anhydrous calcium chloride used in test tube A?
To absorb the moisture in air in the tube
[1]
(c) What is the function of oil used in test tube B?
To prevent the oxygen in air dissolve into the boiled water.
[1]
(d) Suggest two conditions needed for rusting.
Water
[1]
Oxygen
[1]
(e) Draw another set-up to show that the presence of ionic substances can increase the rate
of rusting.
[1]
Compare the degree of rusting with tube C, the nail in this tube forms more
rust.
[1]
(7 marks)
27.
The rusting of iron is investigated by giving five identical iron nails to different treatments. One
nail is left untreated. All six nails are then left for one week exposed to the air. The results are
given in the table below.
Mass of nail and
Mass of nail and
Nail
Treatment
coating before
coating after exposed
exposed to the air
to the air
A Painting
7.0 g
7.3 g
B Coating with wax
7.0 g
6.5 g
C Nickel-plating
7.0 g
7.0 g
D Galvanizing
7.0 g
7.1 g
E Dipping in salt solution
7.0 g
8.9 g
F Untreated
7.0 g
8.2 g
(a) Explain why the mass of the untreated nail has increased after exposed to air.
Because iron reacts with oxygen and water in air to form rust (hydrated
Iron (III) oxide).
[1]
Hydrated iron (III) oxide has increased its mass.
[1]
(b) Which nail is best protected from rusting? Explain your answer.
The nail with nickel-plated;
[1]
Because the iron nail coated with nickel does not increase in mass after
exposed to the air.
[1]
(c) Which treatment make the rusting worse than untreated iron nail? Explain your answer.
(d)
(e)
(f)
Dipping into salt solution;
[1]
The presence of ionic substances speeds up rusting.
[1]
Explain which case is a mistake made in the weighing of the nail.
Treatment B, since the mass of the nail should be the same or higher after
the experiment.
[1]
Which is meant by the treatment 'galvanizing'?
Coated with zinc
[1]
Bridges are protected from rusting by attaching to blocks of magnesium.
(i)
What is the term used to describe this type of rust prevention?
Sacrificial protection
[1]
(ii) Suggest why this method is not used to protect iron nails from rusting?
It is expensive.
[1]
(10 marks)
28.
Aluminium is the most abundant metal on the Earth.
(a) (i)
Name an aluminium ore.
Bauxite
[1]
(ii) Suggest how aluminium can be extracted from its ore.
Electrolysis of molten aluminium oxide.
[1, 1]
(b) One of the special properties of aluminium is that it has high corrosion resistance.
(i)
Explain why aluminium has such a high corrosion resistance.
Since there is a layer of aluminium oxide attached on the metal
surface.
[2]
(ii) The corrosion resistance of aluminium can be further improved by a method.
(1) Name this method.
Aluminium anodization
[1]
(2) Draw the experimental set-up of this method in laboratory.
(c)
[2]
The reaction of aluminium and zinc nitrate solution can be represented by the following
equation:
2Al(s) + 3Zn(NO3)2(aq) → 2Al(NO3)3(aq) + 3Zn(s)
(i)
Give TWO observations of the above reaction.
Aluminium dissolves.
[1]
Grey metal deposits.
[1]
(ii) From the information given, explain why the above reaction is a displacement
reaction.
More reactive aluminium metal displaces less reactive zinc ions from
zinc nitrate solution.
[1]
(iii) If 9 g of aluminium is added into excess zinc nitrate solution, calculate
the
maximum amount of zinc obtained.
(Relative atomic masses: N = 14.0; O = 16.0; Al = 27.0; Zn = 65.4)
Number of moles of aluminium used = 9 / 27.0 = 1/3 mol
Number of moles of zinc formed = (1/3) x (3/2) = 1/2 mol
[1]
Mass of zinc formed = 1/2 x 65.4
[1]
= 32.7 g
[1]
(14 marks)
29.
You are provided with the following materials.
Silver oxide, iron(III) oxide and carbon powder, Bunsen burner, splint, test tubes, delivery tube
and limewater
Suggest three chemical tests to distinguish silver oxide, iron(III) oxide and carbon powder.
(You are required to give a paragraph-length answer. In this question, 6 marks will be awarded
for chemical knowledge and 3 marks for effective communication.)
Heat the three substances alone in air,
[1]
silver oxide will decompose into silvery solid and a gas is released which can
relight a glowing splint.
[1]
Carbon powder will burn to red-hot and a gas is released which is collected with
delivery tube and pass into limewater.
[1]
The gas can turn limewater milky.
[1]
Heat Iron (III) oxide with carbon powder,
[1]
black solid will turn into silvery-grey solid and a gas is released which turns
limewater milky.
[1]
(3 marks for effective communications.)
(9 marks)
30.
You are provided with the following materials.
3 new iron nails, magnesium ribbon, copper foil, gel solution, rust indicator and 3 Petri dishes
Suggest an experiment with explanations which allows you to arrange the reactivity of
magnesium, iron and copper.
(You are required to give a paragraph-length answer. In this question, 6 marks will be awarded
for chemical knowledge and 3 marks for effective communication.)
Wrap two iron nails with magnesium ribbon and copper foil respectively.
[1]
Place an iron nail and two wrapped iron nails into three Petri dishes respectively.
Pour gel solution mixed with rust indicator to each Petri dishes until completely
cover the iron nails.
[1]
Left the dishes for about 20 minutes.
[1]
The reactivity can be determined by observing the presence of blue colour near
the iron nail.
[1]
Since more reactive metal (Mg) can protect iron nail from rusting, no blue colour
will be observed near the iron nail wrapped with Mg
[1]
Less reactive metal (Cu) will speed up the rusting process, so blue colour will be
observed quickly near the iron nail wrapped with Cu
.
[1]
The reactivity order is Cu < Fe < Mg.
(3 marks for effective communications.)
(9 marks)
31.
Protective coatings are commonly used for rust prevention. Discuss the underlying principle
involved and describe TWO kinds of protective coating and their limitations, with reference to
specific examples.
(You are required to give a paragraph-length answer. In this question, 6 marks will be awarded
for chemical knowledge and 3 marks for effective communication.)
Oxygen and water must be present for rusting to occur.
[1]
Protective coatings on iron articles prevent both oxygen and water from reaching
the iron beneath and thus protecting the iron from rusting.
Any two of the followings:
[1]
[2 x 2]
Example 1:
Painting is applied on articles unlikely to be scratched. Cars, bridges and iron
railings are protected from rusting by painting.
However, as soon as the paint is scratched, the metal is exposed to air, and rusting
starts. Therefore painting is not suitable for iron moving parts of machines.
Example 2:
For moving parts of machine, such as door hinge and motor, oil or grease is
applied for rust protection. They also have a lubricating purpose.
However, the oil or grease must be reapplied constantly. Thus this method is
unsuitable for large structures.
Example 3:
Coat hangers and dish racks are often protected with a thin coating of plastic. This
protection lasts longer and improves the appearance of the articles.
However, coating iron articles with plastic is more expensive than painting.
Example 4:
Iron can be coated with a thin layer of tin for rust protection. Food cans are
usually made of mild steel coated with a thin layer of tin because tin ions are
non-poisonous.
However, when the tin coating is damaged, the iron rusts more quickly than when
it is alone. This is because iron is more reactive than tin.
Example 5:
Iron can be coated with a layer of zinc for rust protection. When the zinc coating
is damaged, zinc still protects the iron from rusting. This is because zinc is more
reactive than iron. Therefore zinc corrodes instead of iron. The protection lasts as
long as the zinc lasts.
However, iron coated with zinc is not suitable for making food cans because zinc
ions are poisonous.
Example 6:
We can apply a very thin layer of metal on iron by an electrical process called
electroplating. Chromium and nickel platings protect iron from both oxygen and
water. These platings are not easily scratched off. Chromium-plated articles look
more attractive.
However the cost of chromium-plating is quite high.
(3 marks for effective communications.)
(9 marks)