56
Exercises on stoichiometry: mass, gas volume, and solution volume and
concentration are involved together in these exercises.
Assume that all gas volumes are given at STP (see page 42).
1. Ammonia gas can be produced by heating a mixture of solid sodium hydroxide with
solid ammonium sulfate. What mass of ammonium sulfate should be heated with excess
sodium hydroxide to produce 2.00 L of ammonia gas?
2. Hydrochloric acid is a solution in water of hydrogen chloride gas. What will be the
concentration of the hydrochloric acid if 400 mL of hydrogen chloride are dissolved in
100 mL of water?
3. Will 10.0 mL of 0.20M calcium chloride solution be sufficient to precipitate all of the
silver from 60.0 mL of 0.020M silver nitrate solution?
4. Excess solid zinc carbonate is treated with 40.0 mL of 2.0M hydrochloric acid. What
volume of carbon dioxide is likely to be formed? What will be the concentrations of zinc
and chloride ions in the final solution? (NB zinc chloride is very soluble in water.)
5. Potassium dihydrogenphosphate is going to be made by reacting solid potassium
carbonate with 100.0 mL of 1.00M phosphoric acid (see page 7 for formulae, page 5
for names and formulae of acids). What mass of potassium carbonate is needed?
6. A 25.00mL sample of 0.1012M hydrochloric acid is exactly neutralised by 23.47 mL
of sodium hydroxide solution. It needed 27.33 mL of the same sodium hydroxide
solution to neutralise a second 25.00 mL sample of hydrochloric acid. What is the
concentration of the second sample of hydrochloric acid?
(HINT: there is a short way to solve this problem, that does not require that the
concentration of the sodium hydroxide solution be calculated.)
7. To make 10.0 g of lead iodide, lead nitrate is going to be mixed with 100.0 mL of
0.50M potassium iodide solution. What mass of solid lead nitrate is required? How
much excess potassium iodide is being used?
8. Petrol is mainly octane, C8H18. The mass of 1.00 L of octane is 698 g. What volume of
carbon dioxide, and what mass of water, are formed when 1.00 L of petrol is burned
to form carbon dioxide and water only?
9. When potassium chloride is oxidised, chlorine gas is formed. What volume of chlorine
would be formed if 20.0 g of potassium chloride is oxidised? (HINT: is a full equation
necessary? See page 30).
10. Carbon dioxide and water only are formed when 2.00L of methane gas (CH4) is burnt.
If all of the carbon dioxide formed is passed into a solution containing excess barium
hydroxide, what mass of barium carbonate should be formed?
11. A 20.00 g sample of a mixture of calcium carbonate and calcium sulfate was treated
with excess hydrochloric acid. The carbon dioxide produced was all recovered and its
mass was found to be 3.30 g. What mass of calcium sulfate was present in the sample?
57
ANSWERS TO EXERCISES
Page 15: writing formula:
a) CaO b) LiF c) Na3N d) K2S
e) AlCl3
f) Ca3N2
g) Li2S
h) AlN
i) Al2O3
Page 29: equations for balancing:
Page 34: equations for balancing:
Page 39:
1. 2.0 g
2. 8.0 g
Page 41:
1. 4.9 g
2. C is in excess by 49 g
Page 43:
1. 4.6 g
4. 5.6 L
2. H2 = 1.24 L O2 = 0.62 L
3. NH3 = 6.8 L H2SO4 = 14.8 g
5
5. a) 1.4 x 10 L of oxygen b) 7.0 x 105 L of air.
Page 46:
1. 0.002 mol, 0.15 g
3. 89.2 g
2. 5.74 L
4. 2.3 g
5. 6.7 g
3. No: 13.1 g of zinc are needed
3. 0.162M
4. 958 g
5. 0.38 g
58
Page 47:
1. a) 0.02M b) 2.0M c) 0.40M d) 0.15M e) 0.40M 2. a) 9.3 g b) 17.8 g c) 3.5 g
d) 0.33 g e) 8.1 g 3. a) 0.20M b) 0.20M c) 0.50M d) 0.84M e) 0.091M
4. a) 2.0 L b) 4.0 L
Page 50:
1. 50.0 mL
2. 32.5 mL
3. 0.19M
4. 0.14M
5. 2.5 x 10-3M
6. 10.0 mL
7. 0.020M
The "quick way" mentioned recognises that one mole of dichromate ions produces two
moles of Cr3+ ions. As the solution volume increases from 20 mL to 50 mL, [Cr2O72-]
decreases from 0.025M to 0.010M. However, at the same time, Cr2O72- ions are
changed to Cr3+ ions, so [Cr2O72-] = 0.010M becomes [Cr3+] = 0.020M
8. a) 0.020M b) 0.030M c) 0.0050M d) zero
Explanation: initially, since equal volumes are mixed, the volume doubles so the concentations of the ions is halved, so [Na+] = 0.020M (0.040M in the initial solution
because of two sodium ions in Na2CO3), [CO32-] = 0.010M, [Ca2+] = 0.015M, [NO3-]
= 0.030M (initially doubled like the sodium ions). The amount of calcium ions exceeds
the amount of carbonate ions, so in the reaction all the carbonate precipitates, leaving
0.005 mol L-1 of calcium ions in the solution. Sodium and nitrate ions are spectator
ions, so are unaffected by the reaction.
9. 10.0 mL 10. [Zn2+] = 0.5M, 0.040 mol 11. 13.0 mL 12. [Ba(OH)2] = 3 x [ZnSO4]
13. This problem resembles Example five on page 50. The method of solution suggested
below is a little different from that used on page 50.
Let volume of Ba(OH)2 solution = x
Initial amount of H2SO4 = 20.0 × 0.15 = 3.00 mmol
Amount of barium hydroxide added = 0.10x mmol = amount of sulfuric acid reacted.
Final [H2SO4] = (3.00 - 0.10x) mmol
Final volume of mixture = (20.0 + x) mL
Final [H2SO4] =
= 0.10
This solves to give x = 5.0 mL
Page 56:
1. 5.9 g
2. 0.18M
3. Yes: there are 4 mmol of Cl- to 1.2 mmol of Ag+
4. 0.90 L of CO2, [Zn2+] = 1.0M, [Cl-] = 2.0M
5. 6.91 g
6. 0.1178M
7. 7.2 g of lead nitrate; excess KI = 0.0067 mol or 13 mL of the solution used.
8. Volume of CO2 = 1097 L at STP, mass of H2O = 992 g
9. 3.0 L Explanation about use of a half-equation: the question does not say what oxidises
the chloride, so only a half equation can be written. Potassium is a spectator ion, and
must be included in the equation since the question involves a mass of KCl. (It is
assumed in the equation that KCl is a solid, although the question does not say so.)
10. 17.6 g
11. 12.49 g of CaSO4