Assignment
MOLES AND STOICHIOMETRY
MCQ Questions
1. One mole of each of the following compounds is burnt in excess oxygen. Which compound will
produce three moles of carbon dioxide and three moles of steam only?
a. C3H8
b. C3H7OH
c. C3H7CO2H
d. CH3CO2CH3
2. 20 cm3 of carbon monoxide are reacted with 10 cm3 of oxygen. The equation for the reaction is
shown:
2CO + O2 --> 2CO2
Which volume of carbon dioxide will be produced?
a. 10 cm3
b. 20 cm3
c. 30 cm3
d. 40 cm3
3. What has a mass equal to that of one mole of water?
a. 24 dm3 of water
b. one mole of steam
c. one molecule of water
d. two moles of hydrogen molecules and one mole of hydrogen molecules
4. An 8 g sample of oxygen contains the same number of atoms as 16 g of element X. What is the
relative atomic mass, Ar of X?
a. 4
b. 8
c. 16
d. 32
5. Which quantity is the same for one mole of ethanol and one mole of ethane?
a. mass
b. number of atoms
c. number of molecules
d. volume at r.t.p.
6. Which ion is present in the highest concentration in a 2 mol/dm3 aqueous solution of sodium
sulphate?
a. hydrogen ions
b, hydroxide ions
c. sodium ion
d. sulphate ion
7. Magnesium reacts with hydrochloric acid. Which solution would give the fastest initial rate of
reaction?
a. 40 g of HCl in 1000 cm3 of water
b. 20 g of HCl in 1000 cm3 of water
c. 10 g of HCl in 100 cm3 of water
d. 4 g of HCl in 50 cm3 of water
8. One mole of hydrated copper(II) sulphate, CuSO4.5H2O is dissolved in water. How many moles
of ions does the solution contain?
a. 1
b. 2
c. 6
d. 7
9. What is the ratio of the volume of 2 g of hydrogen to the volume of 16 g of methane, both
volumes at r.t.p.?
a. 1 to 1
b. 1 to 2
c. 1 to 8
d. 2 to 1
10. Which of the following contains the same number of molecules as 9 g of water?
a. 2 g of hydrogen gas
b. 14 g of nitrogen gas
c. 32 g of oxygen gas
d. 44 g of carbon dioxide gas
11. Calcium reacts with water as shown.
Ca (s) + 2H2O (l) --> Ca(OH)2 (aq) + H2 (g)
What is the total mass of the solution that remains when 40g of calcium reacts with 100 g of
water?
a. 58 g
b. 74 g
c. 138 g
d. 140 g
12. 20 cm3 of oxygen are reacted with 20 cm3 of carbon monoxide. What are the volumes of the
gases remaining, at the original temperature and pressure?
oxygen/cm3
carbon
monoxide/cm3
carbon dioxide/cm3
a
0
0
20
b
0
0
40
c
10
0
20
d
10
10
20
13. What is the mass of oxygen contained in 72 g of pure water?
a. 16 g
b. 32 g
c. 64 g
d. 70 g
14. A volume of ethane, C2H6, at r.t.p. has a mass of 20 g. What is the mass of an equal volume of
propene, C3H6 at r.t.p.?
a. 20 g
b. 21 g
c. 28 g
d. 42 g
15. Sodium reacts with water according to the equation below.
2Na + 2H2O --> 2NaOH + H2
Which volume of hydrogen is produced at r.t.p. when 0.2 mol of sodium reacts?
a. 1.2 dm3
b. 2.4 dm3
c. 4.8 dm3
d. 9.6 dm3
16. What is the mass of aluminium in 204g of aluminium oxide, Al 2O3?
a. 26g
b. 27g
c. 54g
d. 108g
17. The equation for the burning of hydrogen is: 2H 2 (g) + O2 (g) ---> 2H2O (g)
One mole of hydrogen gas is made to react with one mole of oxygen gas. What will be present
after the reaction?
a. 1 mol of steam only
b. 1 mol of steam + 0.5 mol of oxygen gas
c. 1 mol of steam + 1 mol of hydrogen gas
d. 2 mol of steam + 0.5 mol of oxygen gas
18. The compound SO2Cl2 reacts with water according to the equation:
SO2Cl2 + 2H2O ---> H2SO4 + 2HCl
How many moles of sodium hydroxide will neutralise the solution produced by one mole
of SO2Cl2 and excess water?
a. 1
b. 2
c. 3
d. 4
19. How many atoms are there in (NH4)SO4?
a. 15
b. 14
c. 10
d. 7
20. 3Cu + 8HNO3 --> XCu(NO3)2 + 2NO + YH2O
What are the appropriate values of X and Y for the balanced reaction equation above?
a. X = 3, Y = 8
b. X = 4, Y = 3
c. X = 3, Y = 4
d. X = 4, Y = 8
21. What is the percentage by mass of calcium in calcium carbonate?
a. 4%
b. 12%
c. 40%
d. 44%
22. Which of the following compounds contain the same percentage by mass of nitrogen as
ammonium cyanate, NH4CNO?
a. NH4NO3
b. NH4Cl2
c. N2H2
d. (NH2)2CO
23. A hydrocarbon contains 20% of carbon and 80% of hydrogen. What is its molecular formula if
it has a relative molecular mass of 30?
a. CH3
b. CH4
c. C2H6
d. C2H4
24. 0.1 mol of metal X (Ar = 27) was burned in oxygen to give an oxide with mass of 5.1g. What is
the formula of the metal oxide if it has a relative molecular mass of 30?
a. XO
b. XO2
c. X2O3
d. X2O
25. 100cm3 of gaseous hydrogen contains n molecules. How many molecules are there in
100cm3 of gaseous methane (CH4) under the same temperature and pressure?
a. n
b. n/5
c. 5n/2
d. 2n/5
26. Which of the following consists of the greatest number of atoms?
a. 4g of H2
b. 8g of O2
c. 71g of Cl2
d. 72dm3 of argon gas at rtp
27. Which of the following consists of the most number of molecules?
a. 32g of O2
b. 18g of water
c. 28g of N2
d. 4g of H2
28. Which of the following formulae for ionic compounds is not correct?
a. Mg(OH)2
b. KF
c. Al2O3
d. Ca2Cl
29. What is the maximum mass of chromium, Cr, that can be extracted from 76g of chromium(III)
oxide?
a. 38g
b. 48g
c. 52g
d. 152g
30. How many oxygen atoms are present in 0.2 moles of N2O5?
a. 6.02 x 1023
b. 4.05 x 1023
c. 3.01 x 1022
d. 6.02 x 1021
MCQ Answers
1. d
2. b
3. b
4. d
5. c (Avogadro's law)
6. c
7. c
8. b
9. a
10. b
11. c (no. of moles of Ca = 40/40 = 1; no. of moles of water = 100/18. So Ca is the limiting reactant and
H2O was present in excess. Mass of hydrogen gas liberated/lost = 2g. Mass of solution = 140 - 2 = 138 g)
12. c
13. c
14. c
15. b
16. d
17. b
18. d
19. a
20. c
21. c
22. d
23. c
24. d
25. a
26. a
27. d
28. d
29. c
30. a
Structured Question Worked Solutions
1. When iron is heated in a steam of dry chlorine, it produces a chloride that contains 34.5% by
mass of iron.
a. calculate the empirical formula of this chloride
b. the relative molecular mass of this chloride (Mr) is 325.
i. what is the molecular formula of this chloride?
ii. hence construct an equation, including state symbols, for the reaction of iron with chlorine.
Solution
1a. molar ratio of Fe : Cl = 34.5/56 : (100 - 34.5)/35.5
= 0.616 : 1.845
=
1
: 3
Hence empirical formula is FeCl3
1bi. let the molecular formula be (FeCl3)n
n x (56 + 3 x 35.5) = 325
--> n = 2
Hence molecular formula is Fe2Cl6
1bii. 2Fe (s) + 3Cl2 (g) --> Fe2Cl6 (s)
2. CFCs are compounds that contain only carbon, chlorine and fluorine. They are atmospheric
pollutants and destroy ozone in the upper atmosphere.
a. 'CFC11' has the following composition by mass
C: 8.7%
F: 13.8%
Cl: 77.5%
Calculate the empirical formula of CFC11
b. 'CFC12' has the molecular formula CF2Cl2. It can be made by the reaction of hydrogen fluoride,
HF, with tetrachloromethane, CCl4
CCl4 + 2HF --> CCl2F2 + 2HCl
What is the maximum mass of CFC12 that can be made from 10.0 g of hydrogen fluoride?
Solution
2a.
C
F
Cl
% mass
8.7
13.8
77.5
Ar
12
19
35.5
molar ratio
8.7/12
= 0.725
=1
13.8/19
= 0.726
=1
77.5/35.5
= 2.183
=3
Empirical formula: CFCl3
2b. Mr of HF = 20
Mr of CF2Cl2 = 121
2 x 20 g of HF give 121 g of CF2Cl2
Therefore, 10 g of HF give (121/40) x 10 = 30.25 g of CF2Cl2
3. Potassium superoxide, KO2 is an ionic solid. It can be used in spacecraft to supply oxygen
according to the following equation.
4KO2 (s) + 2H2O (l) --> 4KOH (s) + 3O2 (g)
The potassium hydroxide formed removes carbon dioxide.
a. show that 1.0 g of potassium superoxide will supply about 0.25 dm 3 of oxygen at room
temperature and pressure.
bi. name the compound formed when carbon dioxide reacts with solid potassium hydroxide
bii. give the equation for the formation of this compound
c. Supplies of oxygen in hospitals are stored in cylinders.
i. state one other use for oxygen
ii. describe briefly how oxygen is obtained from air
Solution
3a. Mr of KO2 = 71
4 x 71 = 284 g of KO2 give 3 x 24 = 72 dm3 of O2 at r.t.p.
hence, 1.0 g of KO2 give (72/284) x 1 = 0.25 dm 3 of O2 at r.t.p
3bi. potassium carbonate
3bii. CO2 (g) + 2KOH (s) --> K2CO3 (s) + H2O (l)
3ci. together with acetylene in welding
3cii. oxygen is obtained by fractional distillation of liquid air
4. Many cars are fitted with air-bags which inflate in an accident. Air-bags contain the solid
sodium azide, NaN3, which decomposes rapidly to form sodium and nitrogen. The nitrogen fills
the air-bag.
a. construct the equation, including state symbols, for the decomposition of sodium azide.
b. in a crash, an air-bag fills with 72 dm3 of nitrogen at room temperature and pressure. What
mass of sodium azide is needed to provide the nitrogen?
c. Sodium azide, NaN3 reacts with dilute hydrochloric acid to give sodium chloride and a
compound A.
Compound A contains 2.33% hydrogen and 97.7% nitrogen by mass.
i. what is the empirical formula of compound A?
ii. construct the equation for the reaction between sodium azide and dilute hydrochloric acid.
Solution
4a. 2NaN3 (s) --> 2Na (s) + 3N2 (g)
4b. no. of moles of nitrogen = 72/24 = 3
no. of moles of NaN3 required = 2
mass of NaN3 = 2 x (23 + 3 x 14) = 130 g
4ci.
element
H
N
%
2.33
97.7
Ar
1
14
No. of moles
2.33
6.98
Ratio
1
3
Empirical formula: N3H
4cii. NaN3 + HCl --> NaCl + N3H
5. Lead white, a white pigment used in old paintings, contains lead(II) carbonate. It darkens when
exposed to air containing traces of hydrogen sulphide, due to the formation of black lead(II)
sulphide, PbS. The white colour can be restored by treating the painting with aqueous hydrogen
peroxide which converts lead(II) sulphide into lead(II) sulphate and water.
ai. Write the chemical equation of the reaction between lead(II) sulphide and hydrogen peoxide.
aii. Calculate the volume of 0.100 mol/dm3 hydrogen peroxide required to react with 0.25 g of
lead(II) sulphide.
b. The element germanium (Ge) was once an important component of transistors. The flow chart
below shows how germanium can be made from its ore germanite.
When 1.00 g of germanite was treated in this way, the germanium present was completely
converted into 0.177 g of a chloride containing 33.9% by mass of germanium.
i. Determine the empirical formula of the chloride.
ii. Write down the valency of germanium in the chloride.
iii. Calculate the percentage of germanium in germanite
Solution
ai. PbS + 4H2O2 ---> PbSO4 + 4H2O
aii. no. of moles of PbS = Mass / Mr
= 0.25 / (207 + 32)
= 0.001046
1 mole of PbS reacts with 4 moles of H2O2
0.001046 moles of PbS react with 0.001046 / 1 x 4 = 0.004184 moles of H2O2
Volume of H2O2 required = no. of moles / concentration
= 0.004184 / 0.100
= 0.0418 dm3
bi.
Elements
Germanite
Chloride
Mass in %
33.9
66.1
Ar
73
35.5
No. of moles
33.9 / 73 = 0.464
66.1 / 35.5 = 1.861
Divide by smallest
number
0.464 / 0.464 = 1
1.861 / 0.464 = 4
--> Empirical formula = GeCl4
bii. valency = 4
biii. Mass of germanium in germanium chloride = 33.9 / 100 x 0.177
= 0.0600 g
% of germanium in germanite = 0.0600 / 1.00 x 100%
= 6.00% (3 sf)
6. Calculate the number of moles of the following:
a. 62.1g of BaCl2
b. H2SO4 in 55cm3 of 0.25 mol/dm3 H2SO4 solution
c. 17dm3 of nitrogen gas N2
Solutions
6a. Mr of BaCl2 = 208
no. of mol of BaCl2 = 62.1/208 = 0.2985 mol --> 0.3mol (2 dp)
6b. no. of mol of H2SO4 = concentration x volume = 0.25 x 55/1000 = 0.01375 --> 0.01 mol (2 dp)
6c. Mr of N2 gas = 28
no. of mol of N2 gas = 17/28 = 0.6071 mol --> 0.61 mol (2 dp)
7. Calculate
a. the concentration of 32g of KOH in 500cm3 of water
b. the mass of iron (Fe) that can be extracted from 14.9g of iron(II) oxide (Fe 2O3)
c. the volume of 100g of CO2 at normal atmospheric pressure and temperature
Solutions
7a. Mr of KOH = 56
no. of mol of KOH = 32/56 = 0.5714mol --> 0.57mol (2 dp)
conc of KOH = no. of mol/(500/1000) = 1.14 mol/dm 3
7b. Mr of Fe2O3 = 160
Mass of Fe extracted = (56 x 2)/160 x 14.9 = 10.43g
7c. Mr of CO2 = 44
no. of mol of CO2 = 100/44 = 2.272 mol --> 2.27 mol (2 dp)
vol of CO2 = 2.27mol x 24dm 3 = 54.48dm3
8. A solution of 250cm3 of ethanol in water contained 22.3g of ethanol, C2H5OH. Calculate the
concentration of ethanol in g/dm3 and mol/dm3.
Solutions
8. Conc of ethanol in g/dm3 = 22.3 / (250/1000) = 89.2 g/dm 3
Mr of ethanol = 46
conc of ethanol in mol/dm3 = 89.2/46 = 1.94 mol/dm3
MORE QUESTIONS
Q1. A mixture of CuSO4–5H2O and MgSO4–7H2O is heated until a mixture of the
anhydrous salt is obtained. If 5.00g of the mixture gives 3.00g of the anhydrous
salts, what is the percentage by mass of CuSO4–5H2O in the mixture?
Solution
CuSO4–5H2O (s)  CuSO4 (s) + 5H2O (s)
Let the mass of CuSO4–5H2O be x g
Number of moles of CuSO4 = Number of moles of CuSO4–5H2O
= x/249.6
MgSO4–7H2O (s)  MgSO4 (s) + 5H2O (s)
Mass of MgSO4–7H2O is (5-x) g
Number of moles of MgSO4 = Number of moles of MgSO4–7H2O
= (5-x)/246.4
Total mass of solid after heating = 3.00g
Mass of CuSO4 + Mass of MgSO4 =3.00g
[ (x/249.6) x 159.6 + ((5-x)/246.4) x 120.4 ] = 3
Solving for x,
x = 3.69
Percentage of CuSO4–5H2O = 3.69/5 x100
= 73.8%
Q2. Solid aluminium sulfide reacts with water to give aluminium hydroxide and
hydrogen sulfide gas. Write a balanced equation for this reaction. What is the
maximum mass of H2S that can form when 158g of aluminium sulfide reacts with
131g of water? Which is the limiting reactant? Calculate the number of moles of
excess reagent remaining at the end of the reaction.
Solution
Al2S3 (s) + 6H2O (l)  2Al(OH)3 (s) + 3H2S (g)
Number of moles of Al2S3 = 158/150.3
= 1.051mol
Number of moles of H2O = 131/18
= 7.28mol
Stoichiometric Coefficient is 1:6:2:3
1.05 x 6 = 6.306
Therefore, Al2S3 is the limiting reactant.
7.28 – 6.306 = 0.974mol
Q3. What is the percentage yield of a reaction in which 41.5g of Tungsten(VI)
Oxide reacts with excess hydrogen gas to produce metallic tungsten and 9.50ml
of water (density = 1.00gml-1)
Solution
WO3 (s) + 3H2 (g)  W (s) + 3H2O (l)
WO3 is the limiting reactant.
Number of moles of WO3 = 41.5/232
= 0.1789mol
Theoretical mass of W formed = 0.1789 x 184
= 32.92g
Actual number of moles of W formed = 1/3 x 9.50/18
= 0.1759mol
Actual mass of W formed = 0.1759 x 184
= 32.37g
Percentage yield of W = 32.37/32.97 x 100%
= 98.3%
Q4. A mixture of 10cm3 of methane and 10cm3 of ethane was sparked with an
excess of oxygen. After cooling to room temperature, the residual gas was
passed through aqueous potassium hydroxide. What volume of gas was
absorbed by the alkali?
Solution
CH4 (s) + 2O2 (g)  CO2 (g) + 2H2O (l)
Stoichiometric Coefficient is 1:2:1:2
Volume of CO2 produced is 10cm3
2C2H6 (g) + 7O2 (g)  4CO2 (g) + 6H2O (l)
Stoichiometric Coefficient is 2:7:4:6
Volume of CO2 produced is 20cm3
Total volume of CO2 produced is 30cm3
Q5. 20cm3 of a gaseous hydrocarbon was mixed with 100cm3 if oxygen so that the
hydrocarbon was completely burnt. The volume of gas remaining at the end of
combustion was 70cm3. After passing over potassium hydroxide, this volume was
reduced to 10cm3. All gases were measured at 25ºC and at the same pressure.
What is the formula of the hydrocarbon?
CXHy + [x+(y/4)] O2  xCO2 + (y/2) H2O (l)
Initial Volume
20cm3
100cm3
0cm3
-
Final Volume
0cm3
10cm3
(70-10)cm3
-
Reacting Volume
20cm3
90cm3
60cm3
-
Mole Ratio
2
9
6
Solution
By comparing coefficients of CxHy and CO2,
2:6 = 1:x
Therefore, x=3
By comparing coefficients of O2 and CxHy,
x+ (y/4) : 1 = 9:2
Because x = 3,
Therefore y =6
Formula of hydrocarbon is C3H6
Q6. 10cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen. A
contraction of 25cm3 in volume occurs after the combustion. On passing the
gaseous products through aqueous sodium hydroxide, a further contraction of
30cm3 occurs. Deduce the formula of the hydrocarbon. (All volumes were
measured at r.t.p.
CXHy + [x+(y/4)] O2  xCO2 + (y/2) H2O (l)
Initial Volume
Final Volume
Reacting Volume
Mole Ratio
Solution
10cm3
0cm3
20cm3
1
Vicm3
Vfcm3
(Vi –Vf ) cm3
?
0cm3
30cm3
60cm3
3
-
Total initial volume – Total final volume = 25cm3
(10 + Vi) – (Vf + 30) = 25cm3
(Vi – Vf) = 45cm3
Mole ratio is 1:4.5:3
By comparing coefficients of CxHy and CO2,
1:3 = 1:x
Therefore, x=3
By comparing coefficients of O2 and CxHy,
x+ (y/4) : 1 = 4.5:1
Because x = 3,
Therefore y =6
Formula of hydrocarbon is C3H6