PS 9.10 Review: Chapter 9 Stoichiometry
1. One of the key steps in producing pure copper from
copper ores is the reaction below:
Cu2S (s) + Cu2O (s) → Cu (s) + SO2 (g) (unbalanced)
After balancing the reaction, determine how many dozen
copper atoms could be produced from a dozen molecules of
the sulfide and two dozen of the oxide of copper(I). Also
how many moles of copper could be produced from one mole
of the sulfide and two moles of the oxide?
a. 1 dozen and 1 mole
b. 2 dozen and 2 moles
c. 3 dozen and 3 moles
d. 6 dozen and 6 moles
2. Carbon tetrachloride, CCl4 (commonly called “carbon
tet”), is used extensively as a solvent in laboratories and in
industry. How many moles of Cl atoms are needed to react
with 2.20 moles of carbon to produce CCl4?
a. 2.20 mole
b. 4.00 moles
c. 4.40 moles
d. 8.80 moles
3. Propane, C3H6, is a common fuel used in heating homes in
rural areas. Predict how many moles of CO2 are formed
when 3.74 moles of propane are burned according to
C3H6 + 6O2  3CO2 + 6H2O
a. 11.2 moles
b. 7.48 moles
c. 3.74 moles
d. 1.25 moles
4. Zirconium is a metal that is used in the design of
catalytic Converters, and that can be isolated from the
carbide in a 2-step process:
ZrC + 4Cl2  ZrCl4 + CCl4
ZrCl4 + 2Mg  Zr + 2MgCl2
How many moles of Zr can be produced from 3.00 moles of
Cl2?
a. 0.25 moles
b. 0.75 moles
c. 1.25 moles
d. 4.00 moles
5. How many grams of oxygen are needed to react with
42.8 g of chromium in the production of Cr2O3, a common
paint pigment?
a. 42.8 g
b. 33.4 g
c. 19.8 g
d. 10.5 g
(p 1 of 2)
6. Ammonium nitrate decomposes when heated to form
“laughing gas,” N2O, and water. If this reaction forms 36.0
g H2O, how many grams of N2O are also formed?
a.
36.0 g
b.
40.0 g
c.
44.0 g
d.
48.0 g
7. Acetylene burners are used in metal welding because
acetylene (C2H2) flames are very hot. How many grams of
calcium carbide are needed to produce 10.0 g C2H2?
CaC2 + 2H2O  Ca(OH)2 + C2H2
a.
24.7 g
b.
64.1 g
c.
10.0 g
d.
15.5 g
8. The Haber process is very important in industry because
it converts atmospheric nitrogen into a more reactive form:
N2(g) + 3H2(g)  2NH3(g)
How many moles of ammonia can theoretically be produced
from a mixture containing 3.0 moles each of nitrogen and
hydrogen?
a.
1.0 mole
b.
2.0 moles
c.
3.0 moles
d.
0.5 moles
9. How many grams of NH3 can be produced from the
mixture of 3.0 g each of nitrogen and hydrogen by the
Haber process?
N2 + 3H2 
2NH3
a.
2.0 g
b.
3.0 g
c.
3.6 g
d.
6.0 g
10. Find the percent yield of product if 1.5 g of SO3 is
produced from 1.0 g of O2 via the reaction:
2S + 3O2  2SO3
a.
75%
b.
80%
c.
88%
d.
92%
PS 9.10 Review: Chapter 9 Stoichiometry
(p 2 of 2)
ANSWERS
1. Choice D becomes obvious once we have balanced the equation:
Cu2S(s)
+
2Cu2O(s)

6Cu(s)
+
SO2(g)
2. D provides the correct response. We need 4 times as many moles of Cl atoms as of C atoms.
3. A provides the correct response:
3.74 moles C3 H 6 x
4. Choice B . Dimensional analysis yields:
3.0 moles Cl2 x
3 moles CO 2
 11.2 moles C3 H 6
1 mole C3 H 6
1 mole Zr
 0.75 moles Zr
4 moles Cl2
5. Choice C can be easily arrived at if we first write the balanced equation and then use dimensional analysis.
4Cr + 3O2

2Cr2O3
1 mole Cr
3 moles O 2
32.0 g O 2
42.8 g Cr x
x
x
 19.8 g Cr
52.0 g Cr
4 moles Cr
1 mole O 2
6. Choice D. First we write a balanced equation and then we use dimensional analysis
NH4NO3

N2O + 2H2O
36.0 g H 2 O x
1 mole H 2 O
1 mole N 2 O
44.0 g N 2 O
x
x
 44.0 g N 2 O
18.0 g H 2 O
2 moles H 2 O
1 mole N 2 O
7. Choice A. Using dimensional analysis:
10.0 g C 2 H 2 x
1 mole C 2 H 2
1 mole CaC2
64.1 g CaC2
x
x
 24.7 g CaC2
26.0 g C 2 H 2
1 mole C 2 H 2
1 mole CaC2
8. Choice B.. First we must realize that this is a limiting reactant problem because specific amounts of both reactants are
provided. Then we must determine which reagent is limiting. In order for all the nitrogen to react, how much hydrogen is
needed?
3 moles H
3.0 moles N 2 x
2
1 mole N 2
 9.0 moles H 2 needed
But we have only 3.0 moles of H2, so H2 is limiting and determines how much product will form:
3.0 moles H 2 x
2 moles NH3
 2.0 mole NH3
3 moles H 2
9. Choice C correctly predicts the amount of product. First we must find the limiting reagent via the number of moles. We
might ask “How much hydrogen is needed to use up all of the nitrogen?”
3.0 g N 2 x
We have :
1 mole N 2
3 moles H 2
x
 0.32 mole H 2 needed
28.0 g N 2
1 mole N 2
3.0 g H 2 x
1 mole H 2
 1.5 moles H 2
2.0 g H 2
This is more than is needed, so there is excess H2 and N2 is limiting.
3.0 g N 2 x
2 moles NH3
17.0 g NH3
1 mole N 2
x
x
 3.6 g NH3
28.0 g N 2
1 mole N 2
1 mole NH3
10. Choice C is correct. First we find the maximum yield of product:
1.0 g O 2 x
2 moles SO 3
80.1 g SO 3
1 mole O 2
x
x
 1.7 g SO 3
32.0 O 2
3 moles O 2
1 mole SO 3
We then divide the yield by the theoretical yield:
1.5 g
x 100%  88%
1.7 g