Name_______________________________
Period_____
Unit 8: Stoichiometry Test Review
Indicate whether the statement is true or false. (1-11)
____
1. Stoichiometry is the study of the relationship between the amount of reactants used and the amount of
products made in a chemical reaction.
____
2. In a balanced chemical equation, the total number of moles of all the reactants is equal to the total number of
moles of all the products.
____
3. In a chemical reaction represented by the general equation
that can be written.
____
4. The mole ratio is a comparison of how many moles of one substance are required to participate in a chemical
reaction with another substance, based on the balanced chemical equation.
____
5. In a chemical reaction, the reactant with the largest molar mass is the limiting reagent.
____
6. The excess reagent refers to the additional amount of a reactant in a chemical reaction that must be added in
order for the reaction to proceed.
____
7. The stoichiometric relationship between any two substances in a reaction depends on the mole ratio between
those substances.
____
8. The limiting reagent limits the amount of product formed in a chemical reaction.
____
9. In a balanced chemical equation, the total number mass of the reactants is equal to the total mass of the
products.
, there are four distinct mole ratios
____ 10. The percent yield is the maximum amount of product that can be produced from a given amount of reactant.
____ 11. If 80.0 g of sodium reacts with 72.0 g of iron (III) oxide, sodium acts as the limiting reactant.
____ 12. Based on the mole ratios of the substances in a chemical reaction shown, determine the correct equation for
the chemical reaction?
Substances
Mole ratio
A:B
3:2
A:C
3:1
B:C
2:1
____ 13. Calculate the mole ratio of all the substances represented in this figure?
____ 14. Which is true of the reaction shown below?
a.
b.
c.
d.
The mole ratio of this reaction is 6:5:6.
Two molecules of Substance Y will be left over when this reaction goes to completion.
Substance Y is the limiting reagent in this reaction.
The addition of more molecules of Substance X will not affect the amount of Substance Z
that can be made.
____ 15. Which conversion factor will correctly complete this setup for finding the number of moles of O2 required to
completely react with 75.0 grams of Sb? The balanced chemical equation for the reaction is shown:
____ 16. The table shows the mole ratios of potassium and bromine combined to form potassium bromide according to
the balanced reaction
. In which trial(s) is the amount of bromine the limiting reagent?
Trial Number
Moles of K
Moles of Br
1
10
3
2
3
2
3
14
8
____ 17. Gold is reacted with chlorine gas according to the reaction 2 Au + 3 Cl2  2 AuCl3. Use the data in the table
to determine the percent yield of gold chloride.
Mass of Gold
Mass of Chlorine
Theoretical Yield
Actual Yield
39.4 g
21.3 g
?
35.2 g
____ 18. According to this chemical reaction, calculate the number of grams of Fe produced from 14 moles of H2?
Fe3O4 (cr) + 4 H2 (g)  3 Fe (cr) + 4 H2O (l)
____ 19. Determine the correct mole ratio of K3PO4 to KNO3 in the chemical reaction
Mg(NO3)2 + K3PO4 Mg3(PO4)2 + KNO3?
____ 20. Determine the correct mole ratio for aluminum chloride to chlorine in the chemical reaction
AlCl3 + Br2  AlBr3 + Cl2?
____ 21. Calculate how many moles of KBr will be produced from 7 moles of BaBr2? BaBr2 + K2SO4  KBr + BaSO4
____ 22. Calculate how many moles of Al would be produced from 20 moles of Al2O3? Al2O3  Al + O2
____ 23. Determine how many moles of Cu are needed to react with 5.8 moles of AgNO3?
Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
____ 24. Which is the number of moles of carbon dioxide produced from the complete combustion of 5.42 moles of
ethanol? C2H6O + O2  CO2 + H2O
____ 25. Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen gas in the presence
of excess ammonia? 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (l)
____ 26. How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?
H2 + O2  H2O
____ 27. How many grams of O2 are required to produce 358.5 grams of ZnO? 2Zn + O2  2ZnO
____ 28. How many grams of bromine are required to react completely with 37.4 grams aluminum chloride?
AlCl3 + Br2  AlBr3 + Cl2
____ 29. How many grams of chlorine gas can be produced from the decomposition of 73.4 g. of AuCl3 by this
reaction: 2AuCl3  2 Au + 3 Cl2
____ 30. A chemical reaction can theoretically produce 137.5 grams of product, but in actuality 112.9 grams are
produced. Which is the percent yield for this reaction?
____ 31. How many moles of carbon dioxide are produced when 19.3 mol of propane gas is burned in excess oxygen?
____ 32. What is the mass of potassium chloride when 6.75 g of potassium reacts with an excess of chlorine gas? The
balanced chemical equation is:
.
____ 33. A certain reaction has a 73.6% yield. If 53.8 grams of the product were predicted by stoichiometry to be
made, what would the actual yield be?
____ 34. A reaction was predicted to produce 32.4 grams of a compound. When the product was measured, there were
only 26.1 grams made. What is the percent yield of this reaction?
____ 35. How many grams of H2O will be produced if 750 grams of Fe are produced?
____ 36. How many grams of Fe3O4 are required to react completely with 300 grams of H2?
____ 37. How many moles of carbon dioxide is produced when 10.4 mol of propane gas is burned in excess oxygen?
____ 38. How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?
____ 39. What is the mass of potassium chloride when 2.50 g of potassium reacts with excess of chlorine gas?
____ 40. How many moles of carbon dioxide are produced when 8.30 mol of ethanol reacts with excess of oxygen?
____ 41. Calculate the mass of citric acid when 2.60 mol of sucrose gas reacts with oxygen.
____ 42. Hydrofluoric acid reacts with 31.3 g of silica to produce hexafluorosilicic acid. Determine the percent yield of
H2SiF6 if the actual yield is 60.3 g.
____ 43. Copper reacts with 36.7 g of silver nitrate to produce copper(II) nitrate and silver. Determine the theoretical
yield of Cu(NO3)2 if the actual yield is 31.29 g.
44. When 200.2 g of sulfur reacts with 100.3 g of chlorine to produce disulfur dichloride,
____________________ acts as the limiting reactant.
45. Nitrogen acts as a(n) ____________________ reactant in the production of ammonia.
46. The mole ratios for the equation A + B  C + D are shown. Based on the mole ratios, write the balanced
general chemical equation represented.
Substances
Mole Ratio
Substances
Mole Ratio
A:B
2:3
B:C
3:1
A:C
2:1
B:D
1:2
A:D
1:3
C:D
1:6
47. Fill in the diagram with the conversion factors that would be used to determine the mass of Substance X
required to form a certain mass of Substance Y.
48. Fill in the boxes with the appropriate number of molecules of Substance Y and Substance Z to show the
decomposition of Substance X in the correct ratios.
49. A dimensional analysis setup to determine the mass of Fe that can be fully oxidized by 8.93 grams of O2 is
shown. Explain what is incorrect in this setup and how to fix the setup to make it correct.
50. Explain why an excess reagent is frequently used in chemical reactions.
51. Find the mole ratio of lead (II) nitrate to potassium nitrate in this chemical reaction.
KI + Pb(NO3)2  PbI2 + KNO3
52. How many moles of zinc chloride can be produced from 15.3 moles of HCl?
Zn + HCl  ZnCl2 + H2
53. In this chemical reaction, 6 moles of each of the reactants is combined.
Mg + N2  Mg3N2
a) Which is the limiting reagent? Which is the excess reagent?
b) How many moles of magnesium nitride can be made?
c) How many moles of excess reagent will be left over?
54. How much chlorine should be produced if 84.2 grams of aluminum chloride and 68.4 grams bromine are
combined?
AlCl3 + Br2  AlBr3 + Cl2
55. 64.9 grams of potassium chloride are reacted with excess oxygen to produce potassium chlorate, according to
the reaction:
KCl + O2 KClO3
If 77.1 grams are produced, what is the percent yield of this reaction?
56. Aluminum chloride and bromine react according to the reaction:
If the percent yield of chlorine is 83.1%, how much aluminum chloride must be used to guarantee a yield of
43.5 grams of chlorine?
57. What is stoichiometry?
58. What is a mole ratio?
59. Balance the following equation and determine the possible mole ratios.
60. List the four steps to solve stoichiometric problems.
61. Titanium dioxide (TiO2) is an industrial chemical used as a white pigment in paint. The conversion of volatile
TiCl4 to TiO2 occurs according to the reaction,
. If 15.70 g of TiCl4
reacts in excess oxygen to form 10.40 g Cl2, what is the percent yield of the reaction?
62. Phosphorus pentachloride is formed when 17.2 g of chlorine gas react with 23.2 g of solid phosphorus (P2).
Determine the reactant that is in excess.
63. In a reaction, 82.00 g of sodium reacts with 74.00 g of ferric oxide to form sodium oxide and iron metal.
Calculate the mass of solid iron produced.
64. In a reaction, 10.76 g of CaCO3, 10.51 g of HCl, and excess water produced 10.26 g of CaCl2 6H2O. Calculate
the theoretical yield of calcium chloride hexahydrate.
65. What is the percent yield for a reaction if the theoretical yield of C6H12 is 21 g and the actual yield recovered
is only 3.8 g?
66. A reaction is predicted to result in 75.0 grams of product being made, but only 58.3 grams are actually
produced. Find the percent yield of this reaction, and explain several reasons why the reaction does not
produce as much as predicted.
67. Balance this equation. Then, show that the Law of Conservation of Matter is being obeyed at the particle
level, the mole level, and the mass level.
Fe2O3 + C  Fe + CO
68. Aluminum metal is burned in oxygen gas.
a. Write the balanced equation for this reaction.
b. If 5.433 g of aluminum is burned with 8.834 g of oxygen gas, what is the limiting reagent?
c. What mass of product can be made?
69. You have 83.6 grams of H2 and 257 grams of N2 which combine according to the following equation:
N2 + 3H2  2NH3
Which reactant is the limiting reagent? Explain how you can tell.
70. 8.344 grams of sodium hydroxide is reacted with 14.290 grams of magnesium nitrate, according to the
reaction NaOH + Mg(NO3)2  NaNO3 + Mg(OH)2.
a) Write the balanced chemical equation for this reaction.
b) Identify the limiting reagent.
c) Determine the theoretical yield of magnesium hydroxide.
71. Represent the reaction between nitrogen and hydrogen in terms of:
a. Particles
b. Moles
c. Mass
72. Represent the reaction between zinc and nitric acid in terms of:
a. Particles
b. Moles
c. Mass
73. In the equation for the combustion of butane, show that the law of conservation of mass is observed. Interpret
the equation for the combustion of butane in terms of:
a. Representative particles
b. Moles
c. Mass
74. A 45.00-g sample of silver nitrate is mixed with 55.00 g of hydrochloric acid to form a white precipitate of
silver chloride. After the solution is filtered and dried, a white precipitate of mass 33.50 g is collected.
a. Determine the limiting reactant.
b. Determine the theoretical yield of silver chloride.
c. Determine the percent yield of silver chloride.
Unit 8: Stoichiometry
Answer Section
TRUE/FALSE
1. ANS: T
PTS: 1
DIF: Bloom's Level 1
NAT: B.3
2. ANS: F
PTS: 1
DIF: Bloom's Level 2
NAT: B.1 | B.2 | B.3
3. ANS: F
PTS: 1
DIF: Bloom's Level 3
NAT: B.1j | B.3
4. ANS: T
PTS: 1
DIF: Bloom's Level 2
NAT: B.1 | B.2 | B.3
5. ANS: F
PTS: 1
DIF: Bloom's Level 2
NAT: B.3
6. ANS: F
PTS: 1
DIF: Bloom's Level 2
NAT: B.3
7. ANS: T
PTS: 1
DIF: Bloom's Level 2
NAT: B.1 | B.2 | B.3
8. ANS: T
PTS: 1
DIF: Bloom's Level 1
NAT: B.3
9. ANS: T
PTS: 1
DIF: Bloom's Level 2
NAT: B.3
10. ANS: F
The percent yield is the ratio of the actual yield to the theoretical yield, and is expressed as a percent.
PTS: 1
DIF: 1
REF: Page 370
OBJ: 12.4.2 Determine the percent yield for a chemical reaction.
NAT: UCP.3 | B.3 TOP: Determine the percent yield for a chemical reaction.
KEY: Theoretical yield
MSC: 1
NOT: The theoretical yield is the maximum amount of product that can be produced from a given amount of
reactant.
11. ANS: F
PTS: 1
DIF: 2
REF: Page 367
OBJ: 12.3.1 Identify the limiting reactant in a chemical equation.
NAT: B.3
TOP: Identify the limiting reactant in a chemical equation.
KEY: Limiting reactant
MSC: 3
NOT: The actual ratio (0.129) is less than the required ratio (0.166). Thus, iron (III) oxide is the limiting
reactant.
MULTIPLE CHOICE
12. ANS: D
NAT: UCP.2 | B.3
PTS: 1
DIF: Bloom's Level 2
13. ANS:
NAT:
14. ANS:
NAT:
15. ANS:
NAT:
16. ANS:
NAT:
17. ANS:
NAT:
18. ANS:
NAT:
19. ANS:
NAT:
20. ANS:
NAT:
21. ANS:
NAT:
22. ANS:
NAT:
23. ANS:
NAT:
24. ANS:
NAT:
25. ANS:
NAT:
26. ANS:
NAT:
27. ANS:
NAT:
28. ANS:
NAT:
29. ANS:
NAT:
30. ANS:
NAT:
31. ANS:
NAT:
32. ANS:
NAT:
33. ANS:
NAT:
34. ANS:
NAT:
35. ANS:
NAT:
36. ANS:
NAT:
37. ANS:
B
PTS:
UCP.2 | B.1 | B.2
B
PTS:
UCP.2 | B.3
C
PTS:
UCP.2 | B.3
D
PTS:
UCP.2 | B.3
C
PTS:
UCP.2 | B.3
A
PTS:
B.3
C
PTS:
B.1 | B.3
B
PTS:
UCP.3 | B.3
C
PTS:
UCP.3
C
PTS:
UCP.3
A
PTS:
UCP.3
C
PTS:
UCP.3 | B.3
A
PTS:
UCP.3 | B.3
C
PTS:
UCP.3 | B.3
D
PTS:
UCP.3 | B.3
D
PTS:
UCP.3 | B.3
A
PTS:
UCP.3 | B.3
A
PTS:
UCP.3 | B.3
B
PTS:
UCP.3 | B.3
C
PTS:
UCP.3 | B.3
D
PTS:
UCP.2 | UCP.3 | B.3
A
PTS:
UCP.3 | B.3
B
PTS:
UCP.3 | B.3
C
PTS:
UCP.3 | B.3
C
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 4
1
DIF: Bloom's Level 4
1
DIF: Bloom's Level 4
1
DIF: Bloom's Level 4
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
1
DIF: Bloom's Level 3
The equation for the combustion of propane is
.
Feedback
A
B
C
D
Divide the unknown moles of carbon dioxide by the known moles of propane.
Multiply the known number of moles of propane by the mole ratio.
Correct!
Balance the equation correctly.
PTS: 1
DIF: 2
REF: Page 359
OBJ: 12.2.2 Use the steps to solve stoichiometric problems.
TOP: Use the steps to solve stoichiometric problems.
KEY: Stoichiometric mole-to-mole conversion
38. ANS: D
The balanced chemical equation is
.
NAT: UCP.1 | UCP.3 | B.3
MSC: 3
Feedback
A
B
C
D
Multiply the number of moles of water by the molar mass of water.
Multiply the number of moles by the mole ratio.
Balance the equation correctly.
Correct!
PTS: 1
DIF: 3
REF: Page 360
OBJ: 12.2.2 Use the steps to solve stoichiometric problems.
TOP: Use the steps to solve stoichiometric problems.
KEY: Stoichiometric mole-to-mass conversion
39. ANS: A
The balanced chemical equation is
.
NAT: UCP.1 | UCP.3 | B.3
MSC: 3
Feedback
A
B
C
D
Correct!
Calculate the mass of KCl using the molar mass as a conversion factor.
Balance the equation correctly.
Convert the grams of KCl to moles using the inverse of molar mass as the conversion
factor.
PTS: 1
DIF: 3
REF: Page 361
OBJ: 12.2.2 Use the steps to solve stoichiometric problems.
TOP: Use the steps to solve stoichiometric problems.
KEY: Stoichiometric mass-to-mass conversion
40. ANS: C
The mole ratio of carbon dioxide to ethanol is 2:1.
Feedback
A
B
C
D
Divide the unknown moles by the known moles.
Multiply the known number of moles by the mole ratio.
Correct!
Balance the equation correctly.
PTS: 1
DIF: 1
REF: Page 359
NAT: UCP.1 | UCP.3 | B.3
MSC: 3
OBJ: 12.2.2 Use the steps to solve stoichiometric problems.
TOP: Use the steps to solve stoichiometric problems.
KEY: Stoichiometric mole-to-mole conversion
41. ANS: D
The molar mass of citric acid is 192.044 g.
NAT: UCP.1 | UCP.3 | B.3
MSC: 3
Feedback
A
B
C
D
Divide the unknown moles by the known moles.
Multiply the known number of moles by the mole ratio and the molar mass.
Balance the equation correctly.
Correct!
PTS: 1
DIF: 2
REF: Page 360
OBJ: 12.2.2 Use the steps to solve stoichiometric problems.
TOP: Use the steps to solve stoichiometric problems.
KEY: Stoichiometric mole-to-mass conversion
42. ANS: D
Percent yield (actual yield/theoretical yield)  100
NAT: UCP.1 | UCP.3 | B.3
MSC: 3
Feedback
A
B
C
D
Multiply the yield by 100 to calculate the percent yield.
Divide the actual yield by the theoretical yield.
The molar mass is incorrect.
Correct!
PTS: 1
DIF: 3
REF: Page 371
OBJ: 12.4.2 Determine the percent yield for a chemical reaction.
NAT: UCP.3 | B.3 TOP: Determine the percent yield for a chemical reaction.
KEY: Percent yield
MSC: 3
43. ANS: D
Percent yield (actual yield/theoretical yield)  100
Feedback
A
B
C
D
Multiply the number of moles by the molar mass to obtain the theoretical yield.
The mole ratio is incorrect.
The molar mass is incorrect.
Correct!
PTS:
OBJ:
NAT:
KEY:
1
DIF: 3
REF: Page 371
12.4.2 Determine the percent yield for a chemical reaction.
UCP.3 | B.3 TOP: Determine the percent yield for a chemical reaction.
Percent yield
MSC: 3
COMPLETION
44. ANS: chlorine
PTS: 1
DIF: 2
REF: Page 365
OBJ: 12.3.1 Identify the limiting reactant in a chemical equation.
NAT: B.3
TOP: Identify the limiting reactant in a chemical equation.
KEY: Limiting reactant
MSC: 3
45. ANS: excess
PTS:
OBJ:
NAT:
TOP:
KEY:
1
DIF: 2
REF: Page 365
12.3.2 Identify the excess reactant and calculate the amount remaining after the reaction is complete.
UCP.3 | B.3
Identify the excess reactant and calculate the amount remaining after the reaction is complete.
Excess reactant
MSC: 1
SHORT ANSWER
46. ANS:
2A + 3B  C + 6D
PTS: 1
DIF: Bloom's Level 3
47. ANS:
A) molar mass of Substance X
B) mole ratio between substance X and substance Y
C) molar mass of substance Y
NAT: UCP.2 | B.1 | B.3
PTS: 1
DIF: Bloom's Level 2
NAT: UCP.2 | UCP.3 | B.3
48. ANS:
Six molecules of diatomic Substance Y, 4 molecules of Substance Z.
PTS: 1
DIF: Bloom's Level 5
NAT: UCP.2 | B.3
49. ANS:
The mole ratio is reversed; it should be 4 moles of Fe/3 moles O2. This is because the number of moles of
oxygen must be on the bottom of the fraction in order to cancel out the moles of oxygen in the previous
fraction (molar mass of oxygen).
PTS: 1
DIF: Bloom's Level 6
NAT: UCP.2 | UCP.3 | B.3
50. ANS:
An excess reagent will accomplish two things. It will ensure that the reaction goes to completion, since all of
the limiting reagent will be able to be used. It will also help to make the reaction go faster.
PTS: 1
51. ANS:
1:2
DIF: Bloom's Level 2
NAT: UCP.2 | B.3
PTS: 1
DIF: Bloom's Level 3
NAT: B.3
52. ANS:
30.6 moles; HCl and ZnCl2 are in a 1:2 molar ratio.
PTS: 1
DIF: Bloom's Level 3
NAT: UCP.3 | B.3
53. ANS:
a) Magnesium is limiting and nitrogen is excess; they react in a 3:1 molar ratio so there is proportionately less
magnesium.
b) 2 moles of product are formed; the mole ratio for the entire reaction is 3:1:1
c) Since 2 moles of nitrogen must be used, there will be 4 moles left over.
PTS: 1
DIF: Bloom's Level 4
NAT: UCP.2 | B.3
54. ANS:
30.3 grams
The bromine is the limiting reagent; 0.428 moles of chlorine will be produced.
PTS: 1
DIF: Bloom's Level 3
NAT: UCP.3 | B.3
55. ANS:
The theoretical yield would be 106.7 grams. Therefore, the percent yield is (77.1/106.7)x100, or 72.3%.
PTS: 1
DIF: Bloom's Level 3
NAT: UCP.2 | UCP.3 | B.3
56. ANS:
With a percent yield of 83.1% and an actual yield of 43.5 grams, the theoretical yield must be at least (43.5
g)/(0.831) = 52.3 grams of chlorine gas. Using this value in stoichiometry, we find that 52.3 grams of chlorine
requires 66.4 grams of aluminum chloride.
PTS: 1
DIF: Bloom's Level 4
NAT: UCP.2 | UCP.3 | B.3
57. ANS:
Stoichiometry is a study of quantitative relationships between the amounts of reactants used and the products
formed by a chemical reaction.
PTS: 1
DIF: 1
REF: Page 354
OBJ: 12.1.1 Identify the quantitative relationships in a balanced chemical equation.
NAT: UCP.3 | B.3 TOP: Identify the quantitative relationships in a balanced chemical equation.
KEY: Stoichiometry
MSC: 1
58. ANS:
A mole ratio is the ratio between the numbers of moles of any two substances in a balanced chemical
equation.
PTS: 1
DIF: 1
REF: Page 356
OBJ: 12.1.2 Determine the mole ratios from a balanced chemical equation.
NAT: UCP.3 | B.3 TOP: Determine the mole ratios from a balanced chemical equation.
KEY: Mole ratio
MSC: 1
59. ANS:
The balanced chemical equation is:
The possible mole ratios are:
PTS: 1
DIF: 2
REF: Page 356
OBJ: 12.1.2 Determine the mole ratios from a balanced chemical equation.
NAT: UCP.3 | B.3 TOP: Determine the mole ratios from a balanced chemical equation.
KEY: Mole ratio
MSC: 3
60. ANS:
The four steps to solve stoichiometric problems are:
a. Write a balanced chemical equation.
b. Determine the moles of a given substance using a mass-to-mole conversion.
c. Determine the moles of an unknown substance from the moles of the given substance.
d. Determine the mass of an unknown substance from the moles of the unknown substance using a mole-tomass conversion.
PTS:
OBJ:
NAT:
TOP:
KEY:
1
DIF: 1
REF: Page 363
12.2.1 Explain the sequence of steps used in solving stoichiometric problems.
UCP.1 | UCP.3 | B.3
Explain the sequence of steps used in solving stoichiometric problems.
Using stoichiometry
MSC: 2
PROBLEM
61. ANS:
88.61%
PTS: 1
DIF: 3
REF: Page 371
OBJ: 12.4.2 Determine the percent yield for a chemical reaction.
NAT: UCP.3 | B.3 TOP: Determine the percent yield for a chemical reaction.
KEY: Percent yield
MSC: 3
NOT: Percent yield = (actual yield/theoretical yield) * 100
62. ANS:
Phosphorus is the reactant that is in excess.
PTS: 1
DIF: 3
REF: Page 367
OBJ: 12.3.2 Identify the excess reactant and calculate the amount remaining after the reaction is complete.
NAT: UCP.3 | B.3
TOP: Identify the excess reactant and calculate the amount remaining after the reaction is complete.
KEY: Excess reactant
MSC: 3
NOT: The actual ratio is less than the required ratio. The number of moles of chlorine needed in the reaction
is 10, but only 1.29 moles of chlorine gas are available. Thus, chlorine acts as the limiting reactant and
phosphorus as the excess reactant.
63. ANS:
51.36 g
PTS: 1
DIF: 3
REF: Page 367
OBJ: 12.3.3 Calculate the mass of a product when the amounts of more than one reactant are given.
NAT: UCP.3 | B.3
TOP: Calculate the mass of a product when the amounts of more than one reactant are given.
KEY: Product mass calculation
MSC: 3
NOT: First, calculate the actual ratio. Then, convert the number of moles of ferric oxide to the number of
moles of iron.
64. ANS:
23.56 g
PTS: 1
DIF: 3
REF: Page 371
OBJ: 12.4.1 Calculate the theoretical yield of a chemical reaction from data.
NAT: UCP.3 | B.3 TOP: Calculate the theoretical yield of a chemical reaction from data.
KEY: Theoretical yield
MSC: 3
NOT: The theoretical yield is calculated by multiplying the number of moles of calcium chloride
hexahydrate by the molar mass.
65. ANS:
18%
PTS:
OBJ:
NAT:
KEY:
NOT:
1
DIF: 3
REF: Page 371
12.4.2 Determine the percent yield for a chemical reaction.
UCP.3 | B.3 TOP: Determine the percent yield for a chemical reaction.
Percent yield
MSC: 3
Percent yield = (actual yield/theoretical yield) * 100
ESSAY
66. ANS:
The percent yield is 77.7%. There may be procedural reasons why the yield is not 100%, such as a precipitate
being left on filter paper or otherwise left behind. Liquids might stick to their containers or evaporate. Finally,
there may be other reactions occurring at the same time, removing some of the reactants and preventing them
from forming the desired product.
PTS: 1
67. ANS:
DIF: Bloom's Level 4
NAT: UCP.3 | B.3
Fe2O3 + 3C  2Fe + 3CO
Particles: 2 atoms Fe, 3 atoms O, 3 atoms C as reactants; 2 atoms Fe, 3 atoms O, 3 atoms C as products
Moles: 2 moles Fe, 3 moles O, 3 moles C as reactants; 2 moles Fe, 3 moles O, 3 moles Cas products
Mass: 1(159.691) + 3(12.01) = 2 (55.847) + 3 (84.027); 195.721 g = 195.721
PTS: 1
DIF: Bloom's Level 6
68. ANS:
a) 4Al(s) + 3O2(g)  2Al2O3(s)
b) The limiting reagent is the aluminum.
c) 10.3 grams of aluminum oxide can be produced.
NAT: UCP.2 | B.3
PTS: 1
69. ANS:
DIF: Bloom's Level 4
NAT: UCP.2 | UCP.3 | B.3
The molar amounts in this reaction are 9.17 moles nitrogen and 41.5 moles hydrogen, based on the molar
masses of each compound. The mole ratio for the entire reaction is 1:3:2. Since 41.5 moles is more than 3
times larger than 9.17 moles, the nitrogen is the limiting reagent. It will only react with 27.5 moles of
hydrogen, leaving 14 moles of hydrogen in excess.
PTS: 1
70. ANS:
DIF: Bloom's Level 4
NAT: UCP.2 | B.3
a) 2 NaOH + Mg(NO3)2  2 NaNO3 + Mg(OH)2.
b) The limiting reagent is NaOH
c) There are 0.1043 moles, or 6.083 grams, of magnesium hydroxide.
PTS: 1
DIF: Bloom's Level 5
71. ANS:
a. 1 molecule N2 + 3 molecules H2  2 molecules NH3
b. 1 mole N2 + 3 moles H2  2 moles NH3
c. 28.02 g N2 + 6.06 g H2  34.08 g NH3
NAT: UCP.2 | B.3
PTS: 1
DIF: 2
REF: Page 355
OBJ: 12.1.2 Determine the mole ratios from a balanced chemical equation.
NAT: UCP.3 | B.3 TOP: Determine the mole ratios from a balanced chemical equation.
KEY: Mole ratio
MSC: 1
72. ANS:
a. 4 atoms Zn + 10 molecules HNO3  4 formula units Zn(NO3)2 + 1 molecule N2O + 5 molecules H2O
b. 4 moles Zn + 10 moles HNO3  4 moles Zn(NO3)2 + 1 mole N2O + 5 moles H2O
c. 261.56 g Zn + 630.2 g HNO3  757.56 g Zn(NO3)2 + 44.02 g N2O + 90.10 g H2O
PTS: 1
DIF: 2
REF: Page 355
OBJ: 12.1.2 Determine the mole ratios from a balanced chemical equation.
NAT: UCP.3 | B.3 TOP: Determine the mole ratios from a balanced chemical equation.
KEY: Mole ratio
MSC: 1
73. ANS:
The equation for the combustion of butane is
.
a. 2 molecules C4H10 + 13 molecules O2  8 molecules CO2 + 10 molecules H2O.
b. 2 moles C4H10 + 13 moles O2  8 moles CO2 + 10 moles H2O.
c. 116 g C4H10 + 416 g O2  352 g CO2 + 180 g H2O.
Mass of reactants Mass of products 532 g.
PTS: 1
DIF: 2
REF: Page 355
OBJ: 12.1.2 Determine the mole ratios from a balanced chemical equation.
NAT: UCP.3 | B.3 TOP: Determine the mole ratios from a balanced chemical equation.
KEY: Mole ratio
MSC: 3
74. ANS:
a. Silver nitrate is the limiting reactant.
b. The theoretical yield of silver chloride is 38.02 g.
c. The percent yield of silver chloride is 88.09%.
PTS:
OBJ:
NAT:
KEY:
1
DIF: 3
REF: Page 371
12.4.2 Determine the percent yield for a chemical reaction.
UCP.3 | B.3 TOP: Determine the percent yield for a chemical reaction.
Percent yield
MSC: 3