Stoich Practice Test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Which branch of chemistry deals with the mass relationships of elements in compounds and the mass
relationships among reactants and products in chemical reactions?
a. qualitative analysis
c. chemical kinetics
b. entropy
d. stoichiometry
____
2. Which of the following would not be studied in the branch of chemistry called stoichiometry?
a. the mole ratio of aluminum and chlorine in aluminum chloride
b. the amount of energy required to break the ionic bonds in calcium fluoride
c. the mass of carbon produced when a known mass of sucrose decomposes
d. the number of moles of hydrogen that reacts completely with a known quantity of oxygen
____
3. The coefficients in a chemical equation represent the
a. masses, in grams, of all reactants and products.
b. relative numbers of moles of reactants and products.
c. number of atoms in each compound in a reaction.
d. number of valence electrons involved in the reaction.
____
4. Each of the four types of reaction stoichiometry problems requires using a
a. table of bond energies.
c. Lewis structure.
b. chart of electron configurations.
d. mole ratio.
____
5. Which equation is not balanced?
a.
b.
____
6. Given the equation
, the starting mass of A, and its molar mass, and you are asked to
determine the moles of C produced, your first step in solving the problem is the multiply the given mass of A
by
a.
c.
b.
____
c.
d.
7. The units of molar mass are
a. g/mol.
b. mol/g.
d.
c. amu/mol.
d. amu/g.
____
8. In the reaction represented by the equation N2 + 3H2  2NH3, what is the mole ratio of nitrogen to ammonia?
a. 1:1
c. 1:3
b. 1:2
d. 2:3
____
9. In the equation 2KClO3  2KCl + 3O2, how many moles of oxygen are produced when 3.0 mol of KClO3
decompose completely?
a. 1.0 mol
c. 3.0 mol
b. 2.5 mol
d. 4.5 mol
____ 10. For the reaction represented by the equation 2H2 + O2 2H2O, how many moles of water can be produced
from 6.0 mol of oxygen?
a. 2.0 mol
c. 12 mol
b. 6.0 mol
d. 18 mol
____ 11. For the reaction represented by the equation AgNO3 + NaCl  NaNO3 + AgCl, how many moles of silver
chloride, AgCl, are produced from 7.0 mol of silver nitrate AgNO3?
a. 1.0 mol
c. 7.0 mol
b. 2.3 mol
d. 21 mol
Use the table below to answer the following questions.
Element
Bromine
Calcium
Carbon
Chlorine
Cobalt
Copper
Fluorine
Hydrogen
Iodine
Iron
Lead
Magnesium
Mercury
Nitrogen
Oxygen
Potassium
Sodium
Sulfur
Symbol
Br
Ca
C
Cl
Co
Cu
F
H
I
Fe
Pb
Mg
Hg
N
O
K
Na
S
Atomic Mass
79.90
40.08
12.01
35.45
58.93
63.55
19.00
1.01
126.90
55.85
207.2
24.30
200.59
14.01
15.00
39.10
22.99
32.01
____ 12. For the reaction represented by the equation 2Na + 2H2O  2NaOH + H2, how many grams of sodium
hydroxide are produced from 3.0 mol of sodium with an excess of water?
a. 40. g
c. 120 g
b. 80. g
d. 240 g
____ 13. For the reaction represented by the equation 2Fe + O2  2FeO, how many grams of iron(II) oxide are
produced from 8.00 mol of iron in an excess of oxygen?
a. 71.8 g
c. 712 g
b. 575 g
d. 1310 g
____ 14. For the reaction represented by the equation 2HNO3 + Mg(OH)2  Mg(NO3)2 + 2H2O, how many grams of
magnesium nitrate are produced from 8.00 mol of nitric acid, HNO3, and an excess of Mg(OH)2?
a. 148 g
c. 593 g
b. 445 g
d. 818 g
____ 15. For the reaction represented by the equation Pb(NO3)2 + 2KI  PbI2 + 2KNO3, how many moles of lead(II)
iodide are produced from 300. g of potassium iodide and an excess of Pb(NO3)2?
a. 0.904 mol
c. 3.61 mol
b. 1.81 mol
d. 11.0 mol
____ 16. For the reaction represented by the equation 3Fe + 4H2O  Fe3O4 + 4H2, how many moles of iron(III) oxide
are produced from 500. g of iron in an excess of H2O?
a. 1.04 mol
c. 8.95 mol
b. 2.98 mol
d. 12.98 mol
____ 17. For the reaction represented by the equation Cl2 + 2KBr  2KCl + Br2, how many grams of potassium
chloride can be produced from 300. g each of chlorine and potassium bromide?
a. 98.7 g
c. 188 g
b. 111 g
d. 451 g
____ 18. For the reaction represented by the equation 2Na + Cl2  2NaCl, how many grams of sodium chloride can be
produced from 500. g each of sodium and chlorine?
a. 112 g
c. 409 g
b. 319 g
d. 824 g
____ 19. Which reactant controls the amount of product formed in a chemical reaction?
a. excess reactant
c. composition reactant
b. mole ratio
d. limiting reactant
____ 20. To determine the limiting reactant in a chemical reaction, one must know the
a. available amount of one of the reactants.
b. amount of product formed.
c. available amount of each reactant.
d. speed of the reaction.
____ 21. To determine the limiting reactant in a chemical reaction involving known masses of A and B, one could first
calculate
a. the mass of 100 mol of A and B.
b. the masses of all products.
c. the bond energies of A and B.
d. the number of moles of B and the number of moles of A available.
____ 22. What is the measured amount of a product obtained from a chemical reaction?
a. mole ratio
c. theoretical yield
b. percentage yield
d. actual yield
____ 23. For the reaction represented by the equation SO3 + H2O  H2SO4, calculate the percentage yield if 500. g of
sulfur trioxide react with excess water to produce 575 g of sulfuric acid.
a. 82.7%
c. 91.2%
b. 88.3%
d. 93.9%
____ 24. For the reaction represented by the equation CH4 + 2O2  2H2O + CO2, calculate the percentage yield of
carbon dioxide if 1000. g of methane react with excess oxygen to produce 2300. g of carbon dioxide.
a. 83.88%
c. 92.76%
b. 89.14%
d. 96.78%
Completion
Complete each statement.
1. If four moles of each reactant are available for the reaction described by the following equation,
____________________ is the substance that is the excess reactant.
Short Answer
1. How does the actual yield of a chemical reaction compare to the theoretical yield?
2. Give at least three reasons why the actual yield of a chemical reaction could be less than the theoretical yield.
Problem
1. Sulfur in gasoline can produce sulfuric acid, H2SO4, according to the two-step process shown below. For each
125 g of sulfur in gasoline, how many moles of H2SO4 will be produced?
2. Given the reaction represented by the equation Mg + 2HCl
H2 + MgCl2, determine to two decimal places
the molar masses of all substances involved. Then write the molar masses as conversion factors.
3. What mass is grams of potassium chloride is produced if 100. g of potassium chlorate decompose according
to the following equation?
4. What mass of PCl3 forms in the reaction of 75.0 g P4 with 275 g Cl2 ?
5. What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal react with excess water
according to the chemical equation 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)?
6. What mass in grams of 1-chloropropane (C3H7Cl) is produced if 400. g of propane react with excess chlorine
gas according to the equation C3H8 + Cl2  C3H7Cl + HCl?
7. What mass in grams of hydrogen gas is produced if 20.0 mol of Zn are added to excess hydrochloric acid
according to the equation Zn(s) +2HCl(aq)  ZnCl2(aq) + H2(g)?
8. How many grams of ammonium sulfate can be produced if 30.0 mol of H2SO4 react with excess NH3
according to the equation 2NH3(aq) + H2SO4(aq)  (NH4)2SO4(aq)?
9. How many moles of Ag can be produced if 350. g of Cu are reacted with excess AgNO3 according to the
equation Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)?
10. The reaction of 100. g of salicylic acid, C7H6O3, with excess acetic anhydride produces 50.0 g of aspirin,
C9H8O4, according to the equation below. What is the percentage yield for this reaction?
11. In the decomposition of hydrogen peroxide, the percentage yield of oxygen is 93.0%. What is the actual yield
in grams of oxygen if you start with 100. g of H2O2? The reaction proceeds according to the equation
.
12. In the reaction represented by the equation 2NH3 + CO2
CO(NH2)2 + H20, 30.7 g of CO(NH2)2 forms per
1.00 mol of CO2 that reacts when NH3 is in excess. What is the percentage yield?
Essay
1. List the four types of reaction stoichiometry problems and briefly describe how to solve each one.
2. Explain the effect of limiting and excess reactants in an automobile engine that stalls.
Stoich Practice Test
Answer Section
MULTIPLE CHOICE
1. ANS: D
PTS: 1
DIF:
REF: cde4a936-f97e-11de-9c72-001185f0d2ea
2. ANS: B
PTS: 1
DIF:
REF: cdebf75d-f97e-11de-9c72-001185f0d2ea
3. ANS: B
PTS: 1
DIF:
REF: cdee32aa-f97e-11de-9c72-001185f0d2ea
4. ANS: D
PTS: 1
DIF:
REF: cdf0bc17-f97e-11de-9c72-001185f0d2ea
5. ANS: D
PTS: 1
DIF:
REF: cdf559c1-f97e-11de-9c72-001185f0d2ea
6. ANS: C
PTS: 1
DIF:
REF: cdfc80d8-f97e-11de-9c72-001185f0d2ea
7. ANS: A
PTS: 1
DIF:
REF: ce016ca2-f97e-11de-9c72-001185f0d2ea
8. ANS: B
PTS: 1
DIF:
REF: ce060a4c-f97e-11de-9c72-001185f0d2ea
9. ANS: D
PTS: 1
DIF:
REF: ce0d3163-f97e-11de-9c72-001185f0d2ea
10. ANS: C
PTS: 1
DIF:
REF: ce0f93c0-f97e-11de-9c72-001185f0d2ea
11. ANS: C
PTS: 1
DIF:
REF: ce121d2d-f97e-11de-9c72-001185f0d2ea
12. ANS: C
Solution:
I
OBJ: 1
II
OBJ: 1
I
OBJ: 1
I
OBJ: 2
II
OBJ: 2
I
OBJ: 2
I
OBJ: 3
III
OBJ: 3
III
OBJ: 3
III
OBJ: 3
III
OBJ: 1
PTS: 1
OBJ: 2
13. ANS: B
Solution:
DIF: III
REF: ce16bad7-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
14. ANS: C
Solution:
DIF: III
REF: ce191d34-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
15. ANS: A
Solution:
DIF: III
REF: ce1b7f91-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 3
16. ANS: B
Solution:
DIF: III
REF: ce1e08fe-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 3
17. ANS: C
Solution:
DIF: III
REF: ce22a6a8-f97e-11de-9c72-001185f0d2ea
Since Cl2 would produce the most KCl, KBr is the limiting reactant, thus 188 G KCl is produced.
PTS: 1
OBJ: 4
18. ANS: D
Solution:
DIF: III
REF: ce276b62-f97e-11de-9c72-001185f0d2ea
Since Na would produce the most NaCl, Cl2 is the limiting reactant, thus 824 g NaCl is produced.
19.
20.
21.
22.
23.
PTS: 1
DIF: III
REF:
OBJ: 4
ANS: D
PTS: 1
DIF:
REF: ce2c572c-f97e-11de-9c72-001185f0d2ea
ANS: C
PTS: 1
DIF:
REF: ce30f4d6-f97e-11de-9c72-001185f0d2ea
ANS: D
PTS: 1
DIF:
REF: ce335733-f97e-11de-9c72-001185f0d2ea
ANS: D
PTS: 1
DIF:
REF: ce381bed-f97e-11de-9c72-001185f0d2ea
ANS: D
Solution:
ce29cdbf-f97e-11de-9c72-001185f0d2ea
I
OBJ: 1
II
OBJ: 1
I
OBJ: 1
I
OBJ: 3
PTS: 1
OBJ: 4
24. ANS: A
Solution:
DIF: III
REF: ce3f4304-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 4
DIF: III
REF: ce41cc71-f97e-11de-9c72-001185f0d2ea
DIF: I
REF: ce4db842-f97e-11de-9c72-001185f0d2ea
COMPLETION
1. ANS:
silicon dioxide (SiO2)
silicon dioxide
SiO2
PTS: 1
OBJ: 1
SHORT ANSWER
1. ANS:
The actual yield is always less than the theoretical yield.
PTS: 1
DIF: II
REF: ce658fe4-f97e-11de-9c72-001185f0d2ea
OBJ: 3
2. ANS:
The actual yield could be less than a theoretical yield for these reasons: reactants may form by-products in
competing side reactions, reactants may contain impurities, and reactions may not go to completion.
PTS: 1
OBJ: 3
DIF: II
REF: ce6a2d8e-f97e-11de-9c72-001185f0d2ea
PROBLEM
1. ANS:
3.90 mol H2SO4
PTS: 1
DIF: III
REF: ce6a549e-f97e-11de-9c72-001185f0d2ea
OBJ: 2
2. ANS:
Mg: 24.30 g/mol; HCl: 36.46 g/mol; H2 : 2.02 g/mol; MgCl2.
95.20 g/mol.
Conversion factors:
PTS: 1
OBJ: 4
3. ANS:
60.8 g KCl
DIF: II
PTS: 1
DIF: III
OBJ: 4
4. ANS:
Assuming that P4 is the limiting reagent:
REF: ce6c8feb-f97e-11de-9c72-001185f0d2ea
REF: ce6ef248-f97e-11de-9c72-001185f0d2ea
Assuming that Cl2 is the limiting reagent:
Since the smaller amount of product is formed from P4, it is the limiting reagent. The mass of product formed
is:
PTS: 1
OBJ: 1
5. ANS:
34.8 g NaOH
DIF: III
REF: ce6f1958-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
6. ANS:
712 g C3H7Cl
DIF: III
REF: ce7154a5-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
7. ANS:
40.4 g H2
DIF: III
REF: ce717bb5-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
8. ANS:
3960 g (NH4)2SO4
DIF: III
REF: ce73b702-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
9. ANS:
11.0 mol Ag
DIF: III
REF: ce76195f-f97e-11de-9c72-001185f0d2ea
DIF: III
REF: ce76406f-f97e-11de-9c72-001185f0d2ea
DIF: II
REF: ce787bbc-f97e-11de-9c72-001185f0d2ea
PTS: 1
OBJ: 2
10. ANS:
PTS: 1
OBJ: 4
11. ANS:
PTS: 1
DIF: II
OBJ: 4
12. ANS:
First find the theoretical yield:
PTS: 1
OBJ: 4
DIF: III
REF: ce7ade19-f97e-11de-9c72-001185f0d2ea
REF: ce7b0529-f97e-11de-9c72-001185f0d2ea
ESSAY
1. ANS:
For all types, start with a balanced chemical equation.
In the first type, both given and unknown quantities are expressed in moles. There is one conversion factor
used to solve it.
In the second type, the given quantity is expressed in moles and the unknown is in grams. Two conversion
factors are needed.
The third type has a given quantity in grams and an unknown amount in moles. This type of problem also
requires two conversion factors.
The last type has both the given and the unknown as masses. Three conversion factors are needed.
PTS: 1
DIF: II
REF: ce7d4076-f97e-11de-9c72-001185f0d2ea
OBJ: 2
2. ANS:
In a properly running engine, air and gasoline are mixed in correct proportions. When air is the limiting
reactant, the engine floods with gas and will not run. When gas is the limiting reactant, the engine stalls
because the amount of fuel burned is not sufficient to run the engine.
PTS: 1
OBJ: 2
DIF: III
REF: ce7fc9e3-f97e-11de-9c72-001185f0d2ea