Chapter 9 Stoichiometry Chapter Review Answers
1.
2.
Mole Ratios
a. Mole ratios are conversion factors that relate the number of moles of one chemical to another chemical involved
in a chemical reaction.
b. They are formed from the coefficients found in a balanced chemical equation.
Mole Ratios as conversion factors
a.
b.
3.
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5.
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2 𝑚𝑜𝑙 𝐶𝑎 1 𝑚𝑜𝑙 𝑂2 2 𝑚𝑜𝑙 𝐶𝑎𝑂
,
,
,
2 𝑚𝑜𝑙 𝐶𝑎
,
1 𝑚𝑜𝑙 𝑂2
,
2 𝑚𝑜𝑙 𝐶𝑎𝑂
. 𝑅𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑎𝑠 𝑤𝑒𝑙𝑙‼
1 𝑚𝑜𝑙 𝑂2 2 𝑚𝑜𝑙 𝐶𝑎 2 𝑚𝑜𝑙 𝐶𝑎 2 𝑚𝑜𝑙 𝐶𝑎𝑂 2 𝑚𝑜𝑙 𝐶𝑎𝑂 1 𝑚𝑜𝑙 𝑂2
1 𝑚𝑜𝑙 𝑀𝑔 2 𝑚𝑜𝑙 𝐻𝐹 1 𝑚𝑜𝑙 𝑀𝑔𝐹2 1 𝑚𝑜𝑙 𝑀𝑔 1 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝑀𝑔
,
,
,
,
,
,
2 𝑚𝑜𝑙 𝐻𝐹
,
1 𝑚𝑜𝑙 𝑀𝑔𝐹2 2 𝑚𝑜𝑙 𝐻𝐹 1 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝑀𝑔𝐹2
,
,
,
2 𝑚𝑜𝑙 𝐻𝐹 1 𝑚𝑜𝑙 𝑀𝑔 1 𝑚𝑜𝑙 𝑀𝑔 1 𝑚𝑜𝑙 𝑀𝑔𝐹2 1 𝑚𝑜𝑙 𝑀𝑔 1 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝑀𝑔𝐹2 2 𝑚𝑜𝑙 𝐻𝐹 1 𝑚𝑜𝑙 𝐻2 2 𝑚𝑜𝑙 𝐻𝐹 1 𝑚𝑜𝑙 𝐻2
105.99 𝑔 𝑁𝑎2 𝐶𝑂3 1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3 74.10 𝑔 𝐶𝑎(𝑂𝐻)2 1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2 40.00 𝑔 𝑁𝑎𝑂𝐻 1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 100.09 𝑔 𝐶𝑎𝐶𝑂3 1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
,
105.99 𝑔 𝑁𝑎2 𝐶𝑂3
,
1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2
,
74.10 𝑔 𝐶𝑎(𝑂𝐻)2
,
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
,
40.00 𝑔 𝑁𝑎𝑂𝐻
,
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
,
100.09 𝑔 𝐶𝑎𝐶𝑂3
,
1 𝑚𝑜𝑙 𝐻2
1 𝑚𝑜𝑙 𝑀𝑔𝐹2
.
Molar Mass
a. Definition: the mass of one mole of a substance
b. Molar mass is used as a conversion factor in stoiciometric calculations to convert mass of a substance to moles of
the substance or convert moles of a substance to mass of the substance.
Hydrogen + Oxygen  Water
a. 5.0 mol H2
b. 2.5 mol O2
Ethane combustion reaction
a. 15.8 mol O2
b. 9.00 mol CO2 & 13.5 mol H2O
575 g Na & 886 g Cl2
Iron production reaction
a. 75.1 mol CO
b. 50.1 mol Fe & 75.1 mol CO2
87,390 g CO or 8.739 × 104 g CO & 12,610 g H2 or 1.261 × 104 g H2
Lightning reaction of N2 & O2
a. 1,10̅0 g NO2 or 1.10 × 103 g NO2
b. 720. g NO or 7.20 × 102 g NO
CO2 removal in spacecraft
a. 42.04 mol NaOH
b. 21.02 mol Na2CO3 & 21.02 mol H2O
Double replacement AgNO3 & NaBr
a. 463 g NaBr
b. 845 g AgBr
Soda-acid fire extinguisher
a. 3.058 mol NaHCO3
b. 3.058 mol CO2, 1.529 mol Na2SO4, 3.058 mol H2O
Sulfuric acid-sodium hydroxide neutralization reaction
a. H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
b. 37 g H2SO4
c. 53 g Na2SO4 & 14 g H2O
Copper & silver nitrate reaction
a. 0.0104 mol Cu(NO3)2
b. 0.0104 mol Cu & 0.0209 mol AgNO3
Aspirin production
a. 13.5 kg C9H8O4
b. 7.66 kg C4H6O3
c. 4.29 L HC2H3O2
Ideal stoichiometry calculations do not account for factors that can affect the relative amounts of product chemicals
formed in reactions. Ideal stoichiometric calculations are based on the amounts of reactants and products under ideal
conditions. Real stoichiometric calculations account for actual conditions under which the reaction occurs.
The limiting reactant is the chemical in the reaction in least abundance for the reaction; it will be the chemical reactant that
runs out first in the reaction. The limiting reactant limits the amount of chemical product(s) that can be formed. The
excess reactant(s) will not be used up completely in the reaction; they will be leftover.
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19. Yields
a.
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The theoretical yield is the maximum amount of product that can be produced from a given amount of a reactant.
The actual is the measured amount of the chemical product(s) obtained in the laboratory.
b. Actual yield is equal to or (& most usually) less than the theoretical yield.
% yield is the ratio comparison of the actual yield to the theoretical yield multiplied by 100 to make the ratio a percent.
Reactions often do not go to completion and some reactants are leftover. Impurities in the reactants throw off the
theoretical yield. Competing side reactions might occur causing different chemical products to be formed. Some of the
product is lost during the collection process, or when being separated and purified.
Limiting reactants:
a. HCl is the limiting reactant
b. Zn is the limiting reactant
c. Fe(OH)3 is the limiting reactant
Excess reactants:
a. 0.5 mol NaOH
b. 1.0 mol HCl
c. 0.5 mol H2SO4
Theoretical yields:
a. 2.0 mol NaCl & 2.0 mol H2O
b. 2.5 mol ZnCl2 & 2.5 mol H2
c. 2.0 mol Fe2(SO4)3 & 12 mol H2O
Cu & AgNO3 single replacement reaction: Cu + 2 AgNO3  2 Ag + Cu(NO3)2
a. Cu is the limiting reactant
b. 0.50 mol AgNO3
c. 2.50 mol Cu(NO3)2 & 5.00 mol Ag
d. 469 g Cu(NO3)2 & 539 g Ag
Sulfuric acid & aluminum hydroxide neutralization reaction:
3 H 2SO4 + 2 Al(OH)3  Al2(SO4)3 + 6 H2O
a. H2SO4 is the limiting reactant
b. 9.05 g excess Al(OH)3
c. 34.9 g Al2(SO4)3 & 11.0 g H2O
Rocket fuel reaction
a. N2O4 is the limiting reactant and will run out 1st.
b. 2,350. kg H2O
Using the % yield equation:
a. 75.0%
b. 0.90 g
c. 95.0%
d. 1.66 g
93.2 g PCl5
10,200 g HNO3 or 1.02 × 104 g HNO3
72.46%
10.2 g NaHCO3 (baking powder)
7.10 × 102 g CH4 or 710. g CH4
1.75 × 103 g CH4 or 1,750 g CH4
547 L water vapor
3.23 g Au
The chemical equation shows the theoretical yield of product(s) expected under ideal conditions. Since it is rare for a
reaction to meet ideal conditions, the actual yield is usually less than the theoretical yield predicted by the equation.
To ensure that all the magnesium is converted to MgO, pure oxygen should be used instead of air. In air, there are other
chemicals, most notably N2, which could react with the Mg producing magnesium nitride, Mg3N2. Also, the pure O2 should
be in excess.
A % yield greater than 100% is not possible because the actual yield can never be greater than the theoretical yield. A
calculated % yield greater than 100% can only happen due to error. For example, the product might not be completely dry,
there could be contaminant chemicals from the reaction, and there might have been side reactions that occurred producing
more massive products than expected.
Smoldering campfire is not getting enough oxygen to produce the products of complete combustion. Blowing on the
smoldering fire might increase the amount of oxygen, allowing the reaction to continue to completion.