Name: ________________________ Class: ___________________ Date: __________
ID: A
Chapter 9 Exam
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. What is the study of the mass relationships among reactants and products in a chemical reaction?
a. reaction stoichiometry
c. electron configuration
b. composition stoichiometry
d. periodic law
____
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. In most chemical reactions the amount of product obtained is
a. equal to the theoretical yield.
c. more than the theoretical yield.
b. less than the theoretical yield.
d. more than the percentage yield.
____
4. To balance a chemical equation, it may be necessary to adjust the
a. coefficients.
c. formulas of the products.
b. subscripts.
d. number of products.
____
5. For the reaction represented by the equation Cl2 + 2KBr → 2KCl + Br2, calculate the percentage yield if 200. g
of chlorine react with excess potassium bromide to produce 410. g of bromine.
a. 73.4%
c. 91.0%
b. 82.1%
d. 98.9%
____
6. Given the equation 3A + 2B → 2C, 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
2 mol C
1 mol A
a.
c.
3 mol A
molar mass A
3 mol A
molar mass A
b.
d.
2 mol C
1 mol A
____
7. When the limiting reactant in a chemical reaction is completely used, the
a. excess reactants begin combining.
c. reaction speeds up.
b. reaction slows down.
d. reaction stops.
____
8. The Haber process for producing ammonia commercially is represented by the equation N2(g) + 3H2(g) →
2NH3(g). To completely convert 9.0 mol hydrogen gas to ammonia gas, how many moles of nitrogen gas are
required?
a. 1.0 mol
c. 3.0 mol
b. 2.0 mol
d. 6.0 mol
____
9. 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%
1
Name: ________________________
ID: A
____ 10. Which equation is not balanced?
a. 2H 2 + O 2 → 2H 2 O
b. 4H 2 + 2O 2 → 4H 2 O
c.
d.
H2 + H2 + O2 → H2O + H2O
2H 2 + O 2 + H 2 0
____ 11. 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.
____ 12. 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
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
____ 13. 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
____ 14. 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
2
Name: ________________________
ID: A
____ 15. For the reaction represented by the equation 2KClO3 → 2KCl + 3O2, how many moles of potassium chlorate
are required to produce 250. g of oxygen?
a. 2.00 mol
c. 4.97 mol
b. 4.32 mol
d. 5.21 mol
____ 16. 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
____ 17. 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
____ 18. For the reaction represented by the equation 2Na + Cl2 → 2NaCl, how many grams of chlorine gas are required
to react completely with 2.00 mol of sodium?
a. 35.5 g
c. 141.8 g
b. 70.9 g
d. 212.7 g
____ 19. 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
____ 20. For the reaction represented by the equation 2H2 + O2 → 2H2O, how many grams of water are produced from
6.00 mol of hydrogen?
a. 2.00 g
c. 54.0 g
b. 6.00 g
d. 108 g
____ 21. For the reaction represented by the equation Mg + 2HCl → H2 + MgCl2, calculate the percentage yield of
magnesium chloride if 100. g of magnesium react with excess hydrochloric acid to yield 330. g of magnesium
chloride.
a. 71.8%
c. 81.6%
b. 74.3%
d. 84.2%
____ 22. 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
____ 23. For the reaction represented by the equation SO3 + H2O → H2SO4, how many grams of sulfuric acid can be
produced from 200. g of sulfur trioxide and 100. g of water?
a. 100. g
c. 245 g
b. 200. g
d. 285 g
____ 24. The units of molar mass are
a. g/mol.
b. mol/g.
c.
d.
3
amu/mol.
amu/g.
Name: ________________________
ID: A
____ 25. To determine the limiting reactant in a chemical reaction, one must know the
a. available amount of one of the reactants. c. available amount of each reactant.
b. amount of product formed.
d. speed of the reaction.
____ 26. Ozone, O3, is produced by the reaction represented by the following equation:
NO 2 (g) + O 2 (g) → NO(g) + O 3 (g)
What mass of ozone will form from the reaction of 2.0 g of NO2 in a car's exhaust and excess oxygen?
a. 1.1 g O3
c. 2.1 g O3
b. 1.8 g O3
d. 4.2 g O3
____ 27. If the percentage yield is equal to 100%, then
a. the actual yield is greater than the theoretical yield.
b. the actual yield is equal to the theoretical yield.
c. the actual yield is less than the theoretical yield.
d. there was no limiting reactant.
____ 28. A chemist interested in the efficiency of a chemical reaction would calculate the
a. mole ratio.
c. percentage yield.
b. energy released.
d. rate of reaction.
____ 29. 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
____ 30. If one knows the mole ratio of a reactant and product in a chemical reaction, one can
a. estimate the energy released in the reaction.
b. calculate the speed of the reaction.
c. calculate the mass of the product produced from a known mass of reactant.
d. decide whether the reaction is reversible.
____ 31. 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
____ 32. In the reaction represented by the equation C + 2H2 → CH4, what is the mole ratio of hydrogen to methane?
a. 1:1
c. 1:2
b. 2:1
d. 2:4
____ 33. Which coefficients correctly balance the formula NH 4 NO 2 → N 2 + H 2 O?
a. 1,2,2
c. 2,1,1
b. 1,1,2
d. 2,2,2
____ 34. A balanced chemical equation allows one to determine the
a. mole ratio of any two substances in the reaction.
b. energy released in the reaction.
c. electron configuration of all elements in the reaction.
d. mechanism involved in the reaction.
4
Name: ________________________
ID: A
____ 35. 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.
____ 36. A determination of the masses and number of moles of sulfur and oxygen in the compound sulfur dioxide
would be studied in
a. reaction stoichiometry.
c. chemical equilibrium.
b. chemical kinetics.
d. composition stoichiometry.
____ 37. What is the ratio of the actual yield to the theoretical yield, multiplied by 100%?
a. mole ratio
c. Avogadro yield
b. percentage yield
d. excess yield
____ 38. Which reactant controls the amount of product formed in a chemical reaction?
a. excess reactant
c. composition reactant
b. mole ratio
d. limiting reactant
____ 39. 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
____ 40. 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
5
ID: A
Chapter 9 Exam
Answer Section
MULTIPLE CHOICE
1. ANS: A
OBJ: 1
2. ANS: B
OBJ: 1
3. ANS: B
OBJ: 3
4. ANS: A
OBJ: 2
5. ANS: C
Solution:
200. g Cl2 ×
6.
7.
8.
9.
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
II
REF: 1
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
I
REF: 1
1 mol Cl2
1 mol Br 2 159.80 g Br 2
×
×
= 450.78 g Br 2
70.90 g Cl2 1 mol Cl2
1 mol Br 2
percentage yield =
experimental yield
× 100
actual yield
percentage yield =
410 g
× 100 = 90.95 or 91.0%
450.78 g
PTS: 1
ANS: C
OBJ: 2
ANS: D
OBJ: 1
ANS: C
OBJ: 3
ANS: D
Solution:
500. g SO 3 ×
DIF: III
PTS: 1
REF: 3
DIF: I
OBJ: 4
REF: 1
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
III
REF: 1
1 mol SO 3
1 mol H 2 SO 4 98.09 g H 2 SO 4
×
×
= 612.5 g H 2 SO 4
80.07 g SO 3
1 mol SO 3
1 mol H 2 SO 4
percentage yield =
experimental yield
× 100
actual yield
percentage yield =
575 g
× 100 = 93.88 or 93.9%
612.5 g
PTS:
10. ANS:
OBJ:
11. ANS:
OBJ:
1
D
2
B
1
DIF: III
PTS: 1
REF: 3
DIF: II
OBJ: 4
REF: 2
PTS: 1
DIF:
REF: 1
1
I
ID: A
12. ANS: D
OBJ: 3
13. ANS: C
Solution:
PTS: 1
8.00 mol HNO 3 ×
PTS: 1
14. ANS: B
Solution:
PTS: 1
15. ANS: D
Solution:
250. g O 2 ×
DIF:
DIF:
PTS: 1
OBJ: 2
III
REF: 2
OBJ: 2
III
REF: 2
OBJ: 3
III
REF: 2
OBJ: 3
2 mol NaOH 40.00 g NaOH
×
= 120.0 or 120 g NaOH
2 mol Na
1 mol NaOH
DIF:
2.00 mol Na ×
500 g Fe ×
REF: 2
1 mol PbI 2
1 mol KI
×
= 0.9035 or 0.904 mol PbI 2
166.00 g KI
2 mol KI
PTS: 1
18. ANS: B
Solution:
PTS: 1
19. ANS: B
Solution:
III
1 mol O 2
2 mol KClO 3
×
= 5.208 or 5.21 mol KClO 3
32.00 g O 2
3 mol O 2
PTS: 1
17. ANS: C
Solution:
3 mol Na ×
REF: 3
71.85 g FeO
2 mol FeO
×
= 574.8 or 575 g FeO
2 mol Fe
1 mol FeO
DIF:
PTS: 1
16. ANS: A
Solution:
300. g KI ×
I
1 mol Mg (NO 3 ) 2
148.32 g Mg(NO 3 ) 2
×
= 593.28 or 593 g Mg(NO 3 ) 2
2 mol HNO 3
1 mol Mg(NO 3 ) 2
DIF:
8.00 mol Fe ×
DIF:
III
REF: 2
OBJ: 2
1 mol Cl2
70.90 g Cl2
×
= 70.90 or 70.9 g Cl2
2 mol Na
1 mol Cl2
DIF:
III
REF: 2
OBJ: 2
1 mol Fe 2 O 3
1 mol Fe
×
= 2.984 or 2.98 mol CO 2
55.85 g Fe
3 mol Fe
DIF:
III
REF: 2
2
OBJ: 3
ID: A
20. ANS: D
Solution:
6.00 mol H 2 ×
2 mol H 2 O 18.02 g H 2 O
×
= 108.12 or 108 g H 2 O
2 mol H 2
1 mol H 2 O
PTS: 1
21. ANS: D
Solution:
100. g Mg ×
DIF:
III
REF: 2
OBJ: 2
95.20 g MgCl2
1 mol Mg
1 mol Mg
×
×
= 391.77 g MgCl2
24.30 g Mg 1 mol MgCl2
1 mol MgCl2
percentage yield =
experimental yield
× 100
actual yield
percentage yield =
330 g
× 100 = 84.23 or 84.2%
391.77 g
PTS: 1
22. ANS: D
OBJ: 1
23. ANS: C
Solution:
DIF: III
PTS: 1
REF: 3
DIF: I
OBJ: 4
REF: 1
200. g SO 3 ×
1 mol SO 3
1 mol H 2 SO 4 98.09 g H 2 SO 4
×
×
= 245.0 g H 2 SO 4
80.07 g SO 3
1 mol SO 3
1 mol H 2 SO 4
100. g H 2 O ×
1 mol H 2 O
1 mol H 2 SO 4 98.09 g H 2 SO 4
×
×
= 544.3 g H 2 SO 4
18.02 g H 2 O
1 mol H 2 O
1 mol H 2 SO 4
Since H2O would produce the most H2SO4, SO3 is the limiting reactant, thus 245 g H2SO4 is produced.
PTS: 1
24. ANS: A
OBJ: 3
25. ANS: C
OBJ: 1
26. ANS: C
Solution:
2.0 g NO 2 ×
PTS:
27. ANS:
OBJ:
28. ANS:
OBJ:
1
B
3
C
3
DIF: III
PTS: 1
REF: 2
DIF: I
OBJ: 4
REF: 1
PTS: 1
DIF:
REF: 3
II
1 mol NO 2
1 mol O 3
48.00 g O 3
×
×
= 2.086 or 2.1 g O 3
46.01 g NO 2 1 mol NO 2
1 mol O 3
DIF: III
PTS: 1
REF: 2
DIF: I
OBJ: 4
REF: 3
PTS: 1
DIF:
REF: 3
3
I
ID: A
29. ANS: D
Solution:
500. g Na ×
1 mol Na
2 mol NaCl 58.44 g NaCl
×
×
= 1270.99 g NaCl
22.99 g Na
2 mol Na
1 mol NaCl
500. g Cl2 ×
1 mol Cl2 2 mol NaCl 58.44 g NaCl
×
×
= 824.26 g NaCl
70.9 g Cl2
1 mol Cl2
1 mol NaCl
Since Na would produce the most NaCl, Cl2 is the limiting reactant, thus 824 g NaCl is produced.
PTS:
30. ANS:
OBJ:
31. ANS:
OBJ:
32. ANS:
OBJ:
33. ANS:
OBJ:
34. ANS:
OBJ:
35. ANS:
OBJ:
36. ANS:
OBJ:
37. ANS:
OBJ:
38. ANS:
OBJ:
39. ANS:
OBJ:
40. ANS:
OBJ:
1
C
2
C
1
B
3
D
2
A
2
D
1
D
1
B
3
D
1
B
3
D
3
DIF: III
PTS: 1
REF: 2
DIF: I
OBJ: 4
REF: 1
PTS: 1
DIF:
III
REF: 2
PTS: 1
DIF:
III
REF: 1
PTS: 1
DIF:
II
REF: 1
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
III
REF: 1
PTS: 1
DIF:
III
REF: 1
4
Chapter 9 Exam [Answer Strip]
D 10.
_____
B 11.
_____
ID: A
D 15.
_____
C 25.
_____
D 35.
_____
C 26.
_____
A 16.
_____
A
_____
1.
D 36.
_____
B
_____
2.
D 12.
_____
C 17.
_____
B 27.
_____
B 37.
_____
B 18.
_____
B
_____
3.
A
_____
4.
B 19.
_____
C 28.
_____
D 38.
_____
D 29.
_____
B 39.
_____
C
_____
5.
D 40.
_____
D 20.
_____
C 30.
_____
C
_____
6.
D 21.
_____
C 31.
_____
D
_____
7.
D 22.
_____
B 32.
_____
C 13.
_____
C
_____
8.
D 33.
_____
C 23.
_____
B 14.
_____
D
_____
9.
A 34.
_____
A 24.
_____
Name: ________________________ Class: ___________________ Date: __________
ID: B
Chapter 9 Exam
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. To determine the limiting reactant in a chemical reaction, one must know the
a. available amount of one of the reactants. c. available amount of each reactant.
b. amount of product formed.
d. speed of the reaction.
____
2. What is the study of the mass relationships among reactants and products in a chemical reaction?
a. reaction stoichiometry
c. periodic law
b. electron configuration
d. composition stoichiometry
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
____
3. 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. 712 g
c. 575 g
b. 1310 g
d. 71.8 g
____
4. For the reaction represented by the equation 2KClO3 → 2KCl + 3O2, how many moles of potassium chlorate
are required to produce 250. g of oxygen?
a. 5.21 mol
c. 4.97 mol
b. 2.00 mol
d. 4.32 mol
1
Name: ________________________
ID: B
____
5. 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. 593 g
c. 148 g
b. 818 g
d. 445 g
____
6. 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. 11.0 mol
c. 3.61 mol
b. 1.81 mol
d. 0.904 mol
____
7. 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. 80. g
c. 120 g
b. 240 g
d. 40. g
____
8. 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. 8.95 mol
c. 2.98 mol
b. 1.04 mol
d. 12.98 mol
____
9. For the reaction represented by the equation 2Na + Cl2 → 2NaCl, how many grams of chlorine gas are
required to react completely with 2.00 mol of sodium?
a. 35.5 g
c. 212.7 g
b. 141.8 g
d. 70.9 g
____ 10. For the reaction represented by the equation 2H2 + O2 → 2H2O, how many grams of water are produced from
6.00 mol of hydrogen?
a. 6.00 g
c. 54.0 g
b. 108 g
d. 2.00 g
____ 11. The coefficients in a chemical equation represent the
a. masses, in grams, of all reactants and products.
b. number of valence electrons involved in the reaction.
c. number of atoms in each compound in a reaction.
d. relative numbers of moles of reactants and products.
____ 12. A determination of the masses and number of moles of sulfur and oxygen in the compound sulfur dioxide
would be studied in
a. chemical kinetics.
c. chemical equilibrium.
b. composition stoichiometry.
d. reaction stoichiometry.
____ 13. To balance a chemical equation, it may be necessary to adjust the
a. coefficients.
c. number of products.
b. subscripts.
d. formulas of the products.
____ 14. When the limiting reactant in a chemical reaction is completely used, the
a. reaction slows down.
c. reaction speeds up.
b. excess reactants begin combining.
d. reaction stops.
2
Name: ________________________
ID: B
____ 15. For the reaction represented by the equation Mg + 2HCl → H2 + MgCl2, calculate the percentage yield of
magnesium chloride if 100. g of magnesium react with excess hydrochloric acid to yield 330. g of magnesium
chloride.
a. 74.3%
c. 81.6%
b. 71.8%
d. 84.2%
____ 16. For the reaction represented by the equation SO3 + H2O → H2SO4, how many grams of sulfuric acid can be
produced from 200. g of sulfur trioxide and 100. g of water?
a. 100. g
c. 285 g
b. 200. g
d. 245 g
____ 17. 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. stoichiometry
c. entropy
b. chemical kinetics
d. qualitative analysis
____ 18. Ozone, O3, is produced by the reaction represented by the following equation:
NO 2 (g) + O 2 (g) → NO(g) + O 3 (g)
What mass of ozone will form from the reaction of 2.0 g of NO2 in a car's exhaust and excess oxygen?
a. 2.1 g O3
c. 1.1 g O3
b. 1.8 g O3
d. 4.2 g O3
____ 19. A chemist interested in the efficiency of a chemical reaction would calculate the
a. percentage yield.
c. energy released.
b. rate of reaction.
d. mole ratio.
____ 20. For the reaction represented by the equation Cl2 + 2KBr → 2KCl + Br2, calculate the percentage yield if 200.
g of chlorine react with excess potassium bromide to produce 410. g of bromine.
a. 98.9%
c. 73.4%
b. 91.0%
d. 82.1%
____ 21. In the equation 2KClO3 → 2KCl + 3O2, how many moles of oxygen are produced when 3.0 mol of KClO3
decompose completely?
a. 2.5 mol
c. 3.0 mol
b. 1.0 mol
d. 4.5 mol
____ 22. In most chemical reactions the amount of product obtained is
a. equal to the theoretical yield.
c. more than the percentage yield.
b. less than the theoretical yield.
d. more than the theoretical yield.
____ 23. 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. 319 g
c. 112 g
b. 409 g
d. 824 g
____ 24. What is the measured amount of a product obtained from a chemical reaction?
a. mole ratio
c. actual yield
b. percentage yield
d. theoretical yield
3
Name: ________________________
ID: B
____ 25. Given the equation 3A + 2B → 2C, 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
3 mol A
2 mol C
a.
c.
2 mol C
3 mol A
molar mass A
1 mol A
b.
d.
1 mol A
molar mass A
____ 26. The Haber process for producing ammonia commercially is represented by the equation N2(g) + 3H2(g) →
2NH3(g). To completely convert 9.0 mol hydrogen gas to ammonia gas, how many moles of nitrogen gas are
required?
a. 6.0 mol
c. 2.0 mol
b. 1.0 mol
d. 3.0 mol
____ 27. Which of the following would not be studied in the branch of chemistry called stoichiometry?
a. the mass of carbon produced when a known mass of sucrose decomposes
b. the number of moles of hydrogen that reacts completely with a known quantity of oxygen
c. the amount of energy required to break the ionic bonds in calcium fluoride
d. the mole ratio of aluminum and chlorine in aluminum chloride
____ 28. Which equation is not balanced?
a. 2H 2 + O 2 → 2H 2 O
b. 2H 2 + O 2 + H 2 0
c.
d.
4H 2 + 2O 2 → 4H 2 O
H2 + H2 + O2 → H2O + H2O
____ 29. To determine the limiting reactant in a chemical reaction involving known masses of A and B, one could first
calculate
a. the bond energies of A and B.
b. the masses of all products.
c. the mass of 100 mol of A and B.
d. the number of moles of B and the number of moles of A available.
____ 30. If the percentage yield is equal to 100%, then
a. the actual yield is greater than the theoretical yield.
b. the actual yield is equal to the theoretical yield.
c. there was no limiting reactant.
d. the actual yield is less than the theoretical yield.
____ 31. If one knows the mole ratio of a reactant and product in a chemical reaction, one can
a. decide whether the reaction is reversible.
b. estimate the energy released in the reaction.
c. calculate the speed of the reaction.
d. calculate the mass of the product produced from a known mass of reactant.
____ 32. Which reactant controls the amount of product formed in a chemical reaction?
a. excess reactant
c. limiting reactant
b. mole ratio
d. composition reactant
____ 33. In the reaction represented by the equation C + 2H2 → CH4, what is the mole ratio of hydrogen to methane?
a. 1:2
c. 1:1
b. 2:1
d. 2:4
4
Name: ________________________
ID: B
____ 34. 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. 91.2%
c. 93.9%
b. 88.3%
d. 82.7%
____ 35. 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. 21 mol
c. 1.0 mol
b. 7.0 mol
d. 2.3 mol
____ 36. In the reaction represented by the equation N2 + 3H2 → 2NH3, what is the mole ratio of nitrogen to ammonia?
a. 1:3
c. 1:1
b. 2:3
d. 1:2
____ 37. The units of molar mass are
a. g/mol.
b. amu/g.
c.
d.
mol/g.
amu/mol.
____ 38. What is the ratio of the actual yield to the theoretical yield, multiplied by 100%?
a. percentage yield
c. mole ratio
b. Avogadro yield
d. excess yield
____ 39. A balanced chemical equation allows one to determine the
a. electron configuration of all elements in the reaction.
b. energy released in the reaction.
c. mechanism involved in the reaction.
d. mole ratio of any two substances in the reaction.
____ 40. Which coefficients correctly balance the formula NH 4 NO 2 → N 2 + H 2 O?
a. 1,2,2
c. 2,2,2
b. 1,1,2
d. 2,1,1
5
ID: B
Chapter 9 Exam
Answer Section
MULTIPLE CHOICE
1. ANS: C
OBJ: 1
2. ANS: A
OBJ: 1
3. ANS: C
Solution:
8.00 mol Fe ×
PTS: 1
7. ANS: C
Solution:
3 mol Na ×
PTS: 1
8. ANS: C
Solution:
500 g Fe ×
PTS: 1
REF: 3
PTS: 1
DIF:
I
REF: 1
III
REF: 2
OBJ: 2
1 mol O 2
2 mol KClO 3
×
= 5.208 or 5.21 mol KClO 3
32.00 g O 2
3 mol O 2
DIF:
8.00 mol HNO 3 ×
300. g KI ×
II
DIF:
PTS: 1
5. ANS: A
Solution:
PTS: 1
6. ANS: D
Solution:
DIF:
71.85 g FeO
2 mol FeO
×
= 574.8 or 575 g FeO
2 mol Fe
1 mol FeO
PTS: 1
4. ANS: A
Solution:
250. g O 2 ×
PTS: 1
III
REF: 2
OBJ: 3
1 mol Mg (NO 3 ) 2
148.32 g Mg(NO 3 ) 2
×
= 593.28 or 593 g Mg(NO 3 ) 2
2 mol HNO 3
1 mol Mg(NO 3 ) 2
DIF:
III
REF: 2
OBJ: 2
1 mol PbI 2
1 mol KI
×
= 0.9035 or 0.904 mol PbI 2
166.00 g KI
2 mol KI
DIF:
III
REF: 2
OBJ: 3
2 mol NaOH 40.00 g NaOH
×
= 120.0 or 120 g NaOH
2 mol Na
1 mol NaOH
DIF:
III
REF: 2
OBJ: 2
1 mol Fe 2 O 3
1 mol Fe
×
= 2.984 or 2.98 mol CO 2
55.85 g Fe
3 mol Fe
DIF:
III
REF: 2
1
OBJ: 3
ID: B
9. ANS: D
Solution:
2.00 mol Na ×
1 mol Cl2
70.90 g Cl2
×
= 70.90 or 70.9 g Cl2
2 mol Na
1 mol Cl2
PTS: 1
10. ANS: B
Solution:
DIF:
6.00 mol H 2 ×
11.
12.
13.
14.
15.
REF: 2
OBJ: 2
2 mol H 2 O 18.02 g H 2 O
×
= 108.12 or 108 g H 2 O
2 mol H 2
1 mol H 2 O
PTS: 1
ANS: D
OBJ: 1
ANS: B
OBJ: 1
ANS: A
OBJ: 2
ANS: D
OBJ: 1
ANS: D
Solution:
100. g Mg ×
III
DIF: III
PTS: 1
REF: 2
DIF: I
OBJ: 2
REF: 1
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
I
REF: 3
95.20 g MgCl2
1 mol Mg
1 mol Mg
×
×
= 391.77 g MgCl2
24.30 g Mg 1 mol MgCl2
1 mol MgCl2
percentage yield =
experimental yield
× 100
actual yield
percentage yield =
330 g
× 100 = 84.23 or 84.2%
391.77 g
PTS: 1
16. ANS: D
Solution:
DIF:
III
REF: 3
OBJ: 4
200. g SO 3 ×
1 mol SO 3
1 mol H 2 SO 4 98.09 g H 2 SO 4
×
×
= 245.0 g H 2 SO 4
80.07 g SO 3
1 mol SO 3
1 mol H 2 SO 4
100. g H 2 O ×
1 mol H 2 O
1 mol H 2 SO 4 98.09 g H 2 SO 4
×
×
= 544.3 g H 2 SO 4
18.02 g H 2 O
1 mol H 2 O
1 mol H 2 SO 4
Since H2O would produce the most H2SO4, SO3 is the limiting reactant, thus 245 g H2SO4 is produced.
PTS: 1
17. ANS: A
OBJ: 1
DIF: III
PTS: 1
REF: 2
DIF: I
2
OBJ: 4
REF: 1
ID: B
18. ANS: A
Solution:
2.0 g NO 2 ×
1 mol NO 2
1 mol O 3
48.00 g O 3
×
×
= 2.086 or 2.1 g O 3
46.01 g NO 2 1 mol NO 2
1 mol O 3
PTS: 1
19. ANS: A
OBJ: 3
20. ANS: B
Solution:
200. g Cl2 ×
DIF: III
PTS: 1
REF: 2
DIF: I
1 mol Cl2
1 mol Br 2 159.80 g Br 2
×
×
= 450.78 g Br 2
70.90 g Cl2 1 mol Cl2
1 mol Br 2
percentage yield =
experimental yield
× 100
actual yield
percentage yield =
410 g
× 100 = 90.95 or 91.0%
450.78 g
PTS: 1
21. ANS: D
OBJ: 3
22. ANS: B
OBJ: 3
23. ANS: D
Solution:
OBJ: 4
REF: 3
DIF: III
PTS: 1
REF: 3
DIF: III
OBJ: 4
REF: 1
PTS: 1
DIF:
REF: 3
I
500. g Na ×
1 mol Na
2 mol NaCl 58.44 g NaCl
×
×
= 1270.99 g NaCl
22.99 g Na
2 mol Na
1 mol NaCl
500. g Cl2 ×
1 mol Cl2 2 mol NaCl 58.44 g NaCl
×
×
= 824.26 g NaCl
70.9 g Cl2
1 mol Cl2
1 mol NaCl
Since Na would produce the most NaCl, Cl2 is the limiting reactant, thus 824 g NaCl is produced.
PTS:
24. ANS:
OBJ:
25. ANS:
OBJ:
26. ANS:
OBJ:
27. ANS:
OBJ:
28. ANS:
OBJ:
29. ANS:
OBJ:
1
C
3
D
2
D
3
C
1
B
2
D
1
DIF: III
PTS: 1
REF: 2
DIF: I
OBJ: 4
REF: 3
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
III
REF: 1
PTS: 1
DIF:
II
REF: 1
PTS: 1
DIF:
II
REF: 2
PTS: 1
DIF:
I
REF: 3
3
ID: B
30. ANS: B
OBJ: 3
31. ANS: D
OBJ: 2
32. ANS: C
OBJ: 1
33. ANS: B
OBJ: 3
34. ANS: C
Solution:
500. g SO 3 ×
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
III
REF: 1
1 mol SO 3
1 mol H 2 SO 4 98.09 g H 2 SO 4
×
×
= 612.5 g H 2 SO 4
80.07 g SO 3
1 mol SO 3
1 mol H 2 SO 4
percentage yield =
experimental yield
× 100
actual yield
percentage yield =
575 g
× 100 = 93.88 or 93.9%
612.5 g
PTS:
35. ANS:
OBJ:
36. ANS:
OBJ:
37. ANS:
OBJ:
38. ANS:
OBJ:
39. ANS:
OBJ:
40. ANS:
OBJ:
1
B
1
D
3
A
3
A
3
D
2
C
2
DIF: III
PTS: 1
REF: 3
DIF: III
OBJ: 4
REF: 2
PTS: 1
DIF:
III
REF: 1
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
I
REF: 3
PTS: 1
DIF:
I
REF: 1
PTS: 1
DIF:
II
REF: 1
4
Chapter 9 Exam [Answer Strip]
A
_____
5.
ID: B
D 15.
_____
D 25.
_____
B 35.
_____
D
_____
6.
C
_____
1.
C 34.
_____
D 16.
_____
D 26.
_____
A
_____
2.
D 36.
_____
C
_____
7.
A 17.
_____
C 27.
_____
C
_____
8.
A 37.
_____
A 18.
_____
A 38.
_____
B 28.
_____
D
_____
9.
D 39.
_____
A 19.
_____
D 29.
_____
B 10.
_____
B 20.
_____
C 40.
_____
D 11.
_____
D 21.
_____
B 12.
_____
B 30.
_____
B 22.
_____
D 31.
_____
C
_____
3.
D 23.
_____
A 13.
_____
A
_____
4.
C 32.
_____
D 14.
_____
C 24.
_____
B 33.
_____
Chapter 9 Exam [Version Map]
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
MC
A
B
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
2
27
22
13
20
25
14
26
34
28
11
24
5
3
4
6
7
9
8
10
15
17
16
37
1
18
30
19
23
31
35
33
40
39
29
12
38
32
36
21
1