problems is to convert to moles, carry out ratio-based calculations, and finally convert
back to grams, number of particles, and so on as required by the equation.
Canadians in Chemistry
Student Textbook page 246
More information on mercury spills and clean-up efforts can be found at:
http://www.ehs.ufl.edu/General/News/Fall97/Mercury.htm, and
http://www.deq.state.mi.us/ead/p2sect/mercury/links.htm.
Solutions for Practice Problems
Student Textbook page 244
Student Textbook page 246
Student Textbook pages 248–249
See Solutions Manual for solutions to Practice Problems.
Section Review Answers
Student Textbook pages 249–250
1. Chemical equations must be balanced so that the law of conservation of mass is
satisfied. If this law is not satisfied, the relationship between products and reactants
will be incorrect, leading to difficulty in solving stoichiometric problems.
1
2. H2(g) + O2(g) → H2O()
2
Fractional coefficients are acceptable because the coefficients of a balanced chemical
equation can represent the number of moles. If, on the other hand, the coefficients
represent atoms, the use of fractional coefficients is not acceptable as there is no such
thing as a fraction of an atom.
1
3. Na(s) + Cl2(g) → NaCl(s)
2
This reaction does not represent 1.0 g of sodium reacting with 0.50 g of chlorine.
The coefficients of a chemical equation do not represent the mass of the molecule or
atom. The coefficients give information on the relative amounts of the
atoms/molecules taking part in the reaction. The above reaction represents 1 mole of
sodium reacting completely with 12 mole of chlorine, or 6.02 × 1023 sodium atoms
reacting with 3.01 × 1023 chlorine molecules.
4. (a) S + O2 → SO2
(b) S + 32 O2 → SO3
3
2
mol O2
1 mol S
3
mol O2
2
32.1 g S
=
(d) n mol S =
32.07 g/mol
3
mol O
n mol O2
= 21 mol S 2
1.00 mol S
(c)
n mol O2
1 mol S
=
n mol O2 =
1.00 mol S
n mol O2 = 1.50 mol O2
m = 1.50 mol O2 × 16.0 g/mol = 24.0 g O2
5. (a)
mole ratio
(b)
n mol O2
0.500 mol C3H8
=
C3H8(g)
+5O2(g)
3CO2(g)
1
5
3
5 mol O2
1 mol C3H8
n mol O2 = 0.500 mol C3H8 ×
5 mol O2
1 mol C3H8
+4H2O(g)
4
= 2.50 mol O2
Chapter 7 Quantities in Chemical Reactions • MHR
265
UNIT 2 Chapter 7 QUANTITIES IN CHEMICAL REACTIONS
(c)
n mol O2
2.00 mol C3H8
5 mol O2
1 mol C3H8
=
n mol O2 = 2.00 mol C3H8 ×
5 mol O2
1 mol C3H8
= 10.0 mol O2
N = 10.0 mol O2 × 6.02 × 1023 molecules/mol = 6.20 × 1024 molecules O2
1 mol C3H8
mol C3H8
= 3.00
(d)
3 mol CO2
n mol CO2
mol CO2
n mol CO2 = 3.00 mol C3H8 × 13 mol
= 9.00 mol CO2
CH
3
6. (a)
PCl5(s)
mole ratio
given/required
=
H3PO4(aq)
4
1
208.22 g/mol
98.00 g/mol
m
23.5 g
23.5 g H3PO4
98.00 g/mol
n mol PCl5
0.240 mol H3PO4
n mol H3PO4 =
1 mol PCl5
1 mol H3PO4
+4H2O( )
1
molar mass
8
+5HCl(aq)
5
= 0.240 mol H3PO4
mol PCl3
n mol H3PO4 = 0.240 mol H3PO4 × 11mol
= 0.240 mol PCl5
H3PO4
m = 0.240 mol PCl5 × 208.22 g/mol = 49.9 g PCl5
(b)
+ 4H2O( )
PCl5(s)
mole ratio
molar mass
given/required
=
4
208.22 g/mol
18.02 g/mol
3.87 g
N
3.87 g PCl5
=
208.22 g/mol
0.0186 mol PCl5
n mol H2O
n mol PCl5 =
1 mol PCl5
4 mol H2O
1
H3PO4(aq)
1
+ 5HCl(aq)
5
0.0186 mol PCl5
n mol H2O = 0.0186 mol PCl5 ×
4 mol H2O
1 mol PCl5
= 7.43 × 10−2 mol H2O
N = 7.43 × 10−2 mol H2O × 6.02 × 1023 molecules/mol
= 4.48 × 1022 molecules H2O
7. (a) Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
(b) n mol PbI2 =
1 mol PbI2
1 mol Pb(NO3)2
1.43 g PbI2
= 3.01 ×
461 g/mol
−3
× 10 mol PbI2
= 3.01
n mol Pb(NO3)2
10−3 mol PbI2
n mol Pb(NO3)2 = 3.01 × 10−3 mol PbI2 ×
1 mol PbI2
1 mol Pb(NO3)2
= 3.01 × 10−3 mol Pb(NO3)2
m = 3.01 × 10−3 mol Pb(NO3)2 × 331.22 g/mol = 1.03 g Pb(NO3)2
8. (a) Early on in the Apollo 13 mission, an oxygen tank exploded and damaged the
command module of the spacecraft. Due to the extensive damage in the module,
the astronauts were confined to the spacecraft’s lunar module. The module was
not designed to support life for an extended period of time and had limited LiOH
scrubbers. The canisters were overloaded and CO2 built up rapidly in the lunar
module.
(b) This problem was solved by creating a network of hoses between the lunar module
and the command module as the scrubbers in the command module had plenty of
LiOH remaining to react with CO2.
266
MHR • Unit 2 Chemical Quantities
UNIT 2 Chapter 7 QUANTITIES IN CHEMICAL REACTIONS
Section Review Answers
Student Textbook pages 258–259
1. Reactants present in excess amounts do not need to be considered as they will not
affect the quantity of product that is produced. For example, a bicycle frame requires
two tires to function properly. If there were an unlimited amount of tires present, but
only ten bicycle frames available, then only ten bicycles can be built. The excess
amount of tires does not affect the quantity of bicycles that can be produced.
2. (a) Oxygen gas will be the reactant in excess because it is found in the air.
(b) A chemist would not have the gold in excess because it is a precious, expensive,
metal. Since gold is extremely unreactive, there is no great need for this metal in
chemical laboratories.
(c) Ideal reactants used in excess are generally inexpensive and readily available.
3.
mole ratio
molar mass
given/required
3Cu(s)
+8HNO3(aq)
3Cu(NO3)2(aq)
3
8
3
63.55 g/mol
63.02 g/mol
30.01g/mol
57.4 g
165 g
m
+2NO(g)
2
+4H2O( )
4
57.4 g Cu
= 0.903 mol Cu
63.55 g/mol
165 g HNO3
HNO3 = 63.02 g/mol = 2.62 mol HNO3
n mol Cu =
n mol
Amount of NO produced based on:
×2
= 0.602 mol
Cu → n = 0.903 mol
3
2.62 mol × 2
= 0.655 mol
HNO3 → n =
8
The limiting reactant is Cu.
m = 0.602 mol NO × 30.01 g/mol = 18.07 g NO
Therefore, 18.07 g of NO will be produced.
4.
Fe2O3(S)
+ 3CO(g)
2Fe(s)
+3CO2(g)
1
3
2
3
159.7 g/mol
28.01 g/mol
55.85 g/mol
mole ratio
molar mass
given/required
11.5 g
n mol Fe2O3 =
11.5 g Fe2O3
159.7 g/mol
2.63 ×10 molecules
24
m
= 0.072mol Fe2O3
2.63 × 10 molecules CO
n mol CO = 6.02
= 4.37 mol CO
× 1023 molecules/mol
Amount of Fe produced based on:
×2
Fe2O3 → n = 0.072 mol
= 0.144 mol
1
24
×2
CO → n = 4.37 mol
= 2.91 mol
3
Fe2O3 is the limiting reactant.
m = 0.144 mol Fe × 55.85 g/mol = 8.04 g Fe
Therefore, 8.04 g of Fe will be produced.
5.
mole ratio
molar mass
given/required
272
MHR • Unit 2 Chemical Quantities
Fe2(SO4)3(aq)
+6NaOH(aq)
3Na2SO4(aq)
+2Fe(OH)3(s)
1
6
3
2
399.91 g/mol
40.00 g/mol
106.88 g/mol
10.0 g
10.0 g
m
10.0 g Fe2(SO4)3
= 0.0250 mol Fe2(SO4)3
399.91 g/mol
10.0 g NaOH
= 0.250 mol NaOH
40.00 g/mol
n mol Fe2(SO4)3 =
Unit Investigation Prep
Student Textbook page 259
n mol NaOH =
Amount of Fe(OH)3 produced based on:
Fe2(SO4)3 → n = 0.0250 1mol × 2 = 0.0500 mol
×2
NaOH → n = 0.250 mol
= 0.0833 mol
6
The limiting reactant is Fe2(SO4)3 .
m = 0.0500 mol Fe(OH)3 × 106.88 g/mol = 5.34 g Fe(OH)3
Therefore, 5.34 g of Fe(OH)3 will be produced.
6.
+2SO2(g)
CS2( )
+4CO(g)
5
2
1
4
12.01 g/mol
64.07 g/mol
76.15 g/mol
17.5 g
225 g
m
5C(s)
mole ratio
molar mass
given/required
17.5 g C
=
12.01 g/mol
225 g SO2
SO2 = 64.07 g/mol
n mol of C =
1.46 mol C
n mol of
= 3.51 mol SO2
Amount of CS2 produced based on:
×1
= 0.291 mol
C → n = 1.46 mol
5
×1
SO2 → n = 3.51mol
= 1.76 mol
2
The limiting reactant is C.
m = 0.29 mol CS2 × 76.15 g/mol = 22.2 g CS2
Therefore, 22.2 g of CS2 will be produced.
7. (a) Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
(b) The chemist would observe the formation of the Cu(s) precipitate and the change
in Zn(s). Also, the chemist would observe the Zn getting smaller and eventually
“disappearing” as it reacted.
(c) The blue colour of the solution in the beaker had not completely disappeared,
indicating that the Zn is the limiting reactant.
(d) 3.12 g of copper(II) chloride is 0.0232 mol of copper(II) chloride. Because zinc
was the limiting reactant, and because zinc and copper(II) chloride are in a 1:1
ratio, the amount of zinc that was added must be less than 0.0232 mol, or 1.52 g.
A+B→C
Mass of A = xg
Mass of B = yg
Mass of C = zg
x is known
y must be determined
z is determined by analysis of
the product
To determine the amount of B,
an excess of A should be used
to carry out the chemical reaction. This process will still not
guarantee that B is the limiting
reactant in the experiment. To
determine the limiting reactant,
the number of moles of C produced should be calculated
and the mole ratio used to
determine how many moles of
A and B should react. If the
number of moles of A that
should react is less than the
amount of A used initially, then
B will be the limiting reactant,
and the mass of B present initially can be easily calculated.
7.3 Percentage Yield
Student Textbook pages 260–270
When chemical reactions occur, predictions of the quantity of product formed do not
always reflect reality. Often, the amount of product formed will differ from the
stoichiometrically calculated value. Reasons for this difference are discussed in section
7.3. The term theoretical yield is used to describe the expected amount of product from
stoichiometric calculations, while the actual yield is determined from experiment.
Investigation 7-B: Determining the Percentage Yield of a Chemical Reaction on pages
266–267 of the student textbook, is an activity in which students will determine the
theoretical yield and actual yield for the reaction between iron and copper(II) chloride.
The stoichiometric relationship between theoretical yield and actual yield will be used to
determine the percentage yield of the solid copper product. The term percentage purity,
and its applications to industry, will also be explored in this section.
Chapter 7 Quantities in Chemical Reactions • MHR
273
UNIT 2 Chapter 7 QUANTITIES IN CHEMICAL REACTIONS
Assessment and Evaluation
ThoughtLab/
ExpressLab/
Investigation
Investigation 7-B:
Determining the
Percentage Yield of
a Chemical
Reaction,
pages 266–267
Curriculum
Expectations
Assessment
Tools/Techniques
Overall Expectations
■ [QCR V.02] carry out experiments and complete
calculations based on quantitative relationships
in balanced chemical reactions
■
Specific Expectations
Understanding Basic Concepts
■ [QCR 1.01] demonstrate an understanding of
Avogadro’s number, the mole concept, and the
relationship between the mole and molar mass
■ [QCR 1.05] state the quantitative relationships
expressed in a chemical equation (e.g., in
moles, grams, atoms, ions, or molecules)
Rubric for
Investigation 7-B:
Determining the
Percentage Yield
of a Chemical
Reaction (see
“Assessment
and Evaluation”
in the front
matter of
Teacher’s
Resource
CD-ROM)
Achievement
Chart Category
■
■
■
Knowledge/
Understanding
Inquiry
Communication
Learning Skills
■
■
■
Organization
Work habits
Teamwork
Developing Skills of Inquiry and Communication
■ [QCR 2.03] solve problems involving quantity in
moles, number of particles, and mass
■ [QCR 2.07] calculate, for any given reactant or
product in a chemical equation, the
corresponding mass or quantity in moles or
molecules of any other reactant or product
■ [QCR 2.08] solve problems involving percentage
yield and limiting reagents
■ [QCR 2.09] compare, using laboratory results,
the theoretical yield of a reaction (e.g., of steel
wool and copper(II) sulfate solution) to the
actual yield, calculate the percentage yield, and
suggest sources of experimental error
Section Review Answers
Student Textbook page 270
1. The percentage yield of a reaction should be calculated using units of grams because
the actual yield is a measured quantity.
2. C7H6O3(s) + CH3OH() → C8H8O3() + H2O()
Percentage yield = 73.6%
3. The percentage yield of the reaction will be less than expected. An impure sample
will not completely react and therefore the actual yield will be less than the theoretical yield, even if the reaction goes to completion.
4. (a) Prepare two samples of copper of equal mass — one pure, and the other impure.
React each sample of copper with excess silver nitrate. Collect the silver precipitate
using filtration, and determine the final mass of silver from both the impure and
pure samples. You will probably want to repeat the experiment several times if you
have the time. Assuming that all of the copper in both samples reacted completely,
the purity of the copper sample can be determined by first calculating the mass of
Cu that reacted in each sample, and then calculate:
Mass of reacted Cu from impure sample
■ Percentage purity =
× 100%
Mass of reacted Cu from pure sample
(b) It is difficult to address the situation of incomplete reactions of pure copper.
Assume that both the copper in the impure sample and the pure copper sample
will undergo reactions with the same completion rate. Determine the percentage
278
MHR • Unit 2 Chemical Quantities
yield of the pure copper sample. Use the percentage yield from the pure copper
sample to determine the amount of silver that should have been produced in the
impure sample. Calculations can then continue to be worked backwards until a
value for the mass of copper that should have reacted is obtained. From here, the
percentage purity of copper can be calculated by dividing the mass of copper that
should have reacted by the original mass of the impure sample. The assumption
that the copper in the pure and impure sample would proceed with the same
percentage yield might not be a good one because the impurities in the impure
sample could interfere with the reaction.
Chapter 7 Review Answers
Student Textbook pages 271–273
Answers to Knowledge/Understanding Questions
1. The coefficients in a balanced chemical equation provide information on the amount
2.
3.
4.
5.
of atoms and molecules that participate in a reaction. The coefficients can also represent the number of moles of a specific atom or molecule participating in a reaction.
A balanced chemical equation is necessary for stoichiometric calculations in order to
accurately determine the relationship between reactants and products in a reaction. A
balanced chemical equation allows one to predict the amount of products expected
from a known amount of reactants, to calculate the necessary amounts of reactants
required to yield a specific amount of product, and to calculate any other quantitative
relationships between reactants and products. Without a balanced chemical equation,
these calculations will be incorrect.
It is not necessary to determine the limiting reactant before beginning any
stoichiometric calculations when the reactants are present in stoichiometric amounts.
In this case, the reactants are present in a mole ratio that corresponds exactly to the
mole ratio predicted by the balanced chemical equation. That is, when the reaction is
complete, there are no reactants left. Also, when the amount of limiting reactant is
given, and the other reactants are stated to be in excess, there is no need to do any
calculations to determine the limiting reactant. In decomposition reactions, there is
only one reactant present. Therefore, there is no need to determine the limiting
reactant.
The concept of percentage yield was introduced because the theoretical yield
predicted from the balanced chemical equation is not always the same as the actual
yield obtained in an experiment.
The student’s reasoning is incorrect. The coefficients of a chemical equation do not
represent the mass ratio of the molecules or atoms. The coefficients give information
on the relative amounts of the atoms/molecules taking part in the reaction. The
correct way for the student to determine the mass of aluminum oxide produced
would be to state that 4 mol of aluminum reacts with 3 mol of oxygen to produce
2 mol of aluminum oxide. The mass of the product can then be calculated using this
information.
Chapter 7 Quantities in Chemical Reactions • MHR
279
UNIT 2 Chapter 7 QUANTITIES IN CHEMICAL REACTIONS
Answers to Inquiry Questions
6.
mole ratio
molar mass
given/required
4 mol Al
3 mol O2
=
4Al(s)
+3O2(g)
2Al2O3(s)
4
3
2
26.98 g/mol
32.00 g/mol
0.400 mol
m
0.400 mol Al
n mol O2
O2
n mol O2 = 0.400 mol Al × 34 mol
= 0.300 mol O2
mol Al
m = 0.300 mol O2 × 32.00 g/mol = 9.60 g O2
7.
mole ratio
molar mass
given/required
Ca(s)
+Cl2(g)
CaCl2(s)
1
1
1
40.08 g/mol
110.98 g/mol
5.3 g
N
5.3 g Ca
40.08 g/mol
0.13 mol Ca
n mol CaCl2
Amount of Ca =
1 mol Ca
1 mol CaCl2
=
= 0.13 mol Ca
1 mol CaCl2
1 mol Ca
n mol Ca = 0.13 mol Ca ×
= 0.13 mol CaCl2
N = 0.13 mol CaCl2 × 6.02 × 1023 formula units/mol
= 8.0 × 1022 formula units of CaCl2
8.
C3H8
mole ratio
+5O2(g)
1
molar mass
given/required
5
3CO2(g)
3
44.11 g/mol
44.01 g/mol
97.5 g
m
97.5 g C3H8
=
44.11 g/mol
2.21 mol × 3
= 6.63
1
Amount of C3H8 =
+4H2O(g)
4
2.21 mol
n mol CO2 =
mol
m = 6.63 mol CO2 × 44.01g/mol = 291.8 g CO2
9.
Zn(s)
mole ratio
molar mass
given/required
+S(s)
1
1
1
65.39 g/mol
32.07 g/mol
97.46 g/mol
6.00 g
3.35 g
m
6.00 g Zn
= 0.0918 mol
65.39 g/mol
3.35 g S
= 0.104 mol S
32.07 g/mol
(a) n mol Zn =
n mol S =
ZnS(s)
Zn
Amount of ZnS produced based on:
0.0918 mol × 1
1
0.104 mol × 1
=
1
Zn → n =
= 0.0918 mol
S→n=
0.104 mol
The limiting reactant is Zn.
(b) m = 0.0918 mol ZnS × 97.46 g/mol = 8.95 g ZnS
0.0918 mol × 1
= 0.0918 mol
(c) Amount S used in the reaction =
1
Amount of S unused in the reaction = 0.104 mol − 0.0918 mol = 0.012 mol
Mass of S remaining after the reaction = 0.012 mol S × 32.07 g/mol = 0.39 g S
280
MHR • Unit 2 Chemical Quantities
10.
TiCl 4(s)
mole ratio
+2H 2O( )
TiO2(s)
+4HCl(g)
2
1
4
1
molar mass
189.67 g/mol
79.87 g/mol
85.6 g
m
given/required
85.6 g TiCl4
= 0.451 mol
189.67 g/mol
0.451 mol × 1
TiO2 produced =
= 0.451
1
Amount of TiCl4 =
Amount of
mol
m = 0.451 mol TiO2 × 79.87 g/mol = 36.0 g TiO2 produced
11.
4Ag(s)
+2H2S(g)
+O2(g)
2Ag2S(s)
+2H2O(g)
4
2
1
2
2
107.87 g/mol
34.09 g/mol
32.00 g/mol
247.81 g/mol
1.90 g
0.280 g
0.160 g
m
mole ratio
molar mass
given/required
1.90 g Ag
= 0.0176 mol
107.87 g/mol
0.280 g H2S
H2S = 34.09 g/mol = 0.00821 mol
0.160 g O
O2 = 32.00 g/mol2 = 0.00500 mol
Amount of Ag =
Amount of
Amount of
Amount of Ag2S produced based on:
Ag → n = 0.0176 4mol × 2 = 0.00880 mol
0.00821 mol × 2
= 0.00821 mol
2
0.00500 mol × 2
= 0.0100 mol
1
H2S → n =
O2 → n =
The limiting reactant is H2S.
m = 0.00821 mol Ag2S × 247.81 g/mol = 2.03 g Ag2S
12.
2Ca3(PO4)2(s)
mole ratio
+6SiO2(s)
+10C(s)
6
10
2
molar mass
P4(s)
1
+10CO(g)
6
10
310.18 g/mol 60.09 g/mol 12.01 g/mol
116.17 g/mol
13.3 g
m
given/required
20.8 g
3.90 g
20.8 g Ca3(PO4)2
= 0.0671
310.18 g/mol
13.3 g SiO2
SiO2 = 60.09 g/mol = 0.221 mol
3.90 g C
C = 12.01 g/mol = 0.325 mol
Amount of Ca3(PO4)2 =
Amount of
+6CaSiO3(s)
mol
Amount of
Amount of CaSiO3 produced based on:
×6
= 0.201 mol
Ca3(PO4)2 → n = 0.671 mol
2
0.221 mol × 6
6
0.325 mol × 6
=
10
SiO2 → n =
= 0.221 mol
C→n=
0.195 mol
The limiting reactant is C.
m = 0.195 mol CaSiO3 × 116.17 g/mol = 22.7 g CaSiO3
13.
3As2S3(s) +4H2O( ) +10HNO3(aq) +18NaNO3(aq)
mole ratio
molar mass
given/required
3
4
10
246.05 g/mol 18.02 g/mol 63.02 g/mol
1.56 g
Amount of As2S3 =
0.140 g
1.56 g AS2S3
246.05 g/mol
1.23 g
18
9Na2SO4(aq) +6H3AsO4(aq) +28NO(g)
9
6
85.00 g/mol
141.95 g/mol
3.50 g
m
28
= 0.00634 mol
Chapter 7 Quantities in Chemical Reactions • MHR
281
UNIT 2 Chapter 7 QUANTITIES IN CHEMICAL REACTIONS
0.140 g H2O
= 0.00777 mol
18.02 g/mol
1.23 g HNO
HNO3 = 63.02 g/mol3 = 0.0195 mol
3.50 g NaNO
NaNO3 = 85.00 g/mol 3 = 0.0412 mol
Amount of H2O =
Amount of
Amount of
Amount of H3AsO4 produced based on:
As2S3 → n = 0.006343mol × 6 = 0.01268 mol
0.00777 mol × 6
= 0.01166 mol
4
0.0195 mol × 6
HNO3 → n =
= 0.01171 mol
10
NaNO3 → n = 0.041218mol × 6 = 0.01373 mol
H2O → n =
The limiting reactant is H2O.
m = 0.01166 mol H3AsO4 × 141.95 g/mol = 1.66 g H3AsO4
14.
C5H12( )
mole ratio
1
molar mass
72.17 g/mol
8
5
32.00 g/mol
44.01 g/mol
3.00 g
m
2
2.85 × 10 g
given/required
2.85 × 102 g C5H12
= 3.95
72.17 g/mol
3.00 g O2
= 0.0938 mol
32.00 g/mol
Amount of C5H12 =
+6H2O( )
5CO2(g)
+8O2(g)
6
mol
Amount of O2 =
Amount of CO2 produced based on:
×5
= 19.75 mol
C5H12 → n = 3.95 mol
1
O2 → n = 0.0938 8mol × 5 = 0.0586 mol
The limiting reagent is O2 .
m = 0.0586 mol CO2 × 44.01 g/mol = 2.58 g CO2
15.
SiO2(s)
mole ratio
1
molar mass
60.01 g/mol
given/required
+4HF(aq)
SiF4(g)
+2H2O(g)
4
1
2
104.09 g/mol
18.02 g/mol
m
12.2 g
(a) Amount of of SiO2 =
12.2 g SiO2
60.01 g/mol
= 0.203 mol
×2
Amount of of H2O produced = 0.203 mol
= 0.406 mol
1
Theoretical yield of H2O = 0.406 mol H2O × 18.02 g/mol = 7.32 g H2O
2.50 g
× 100% = 34.2%
(b) Percentage yield =
7.32 g
(c) Amount of of SiF4 produced =
0.203 mol × 1
1
= 0.203 mol
Theoretical yield of SiF4 = 0.203 mol SiF4 × 104.09 g/mol = 21.1 g SiF4
Assuming a 34.2% yield,
the actual mass of SiF4 = 21.13 g SiF4 × 0.342 = 7.23 g SiF4
16.
mole ratio
molar mass
given
BaCl2(s)
+Na2SO4(aq)
BaSO4(s)
+2NaCl(aq)
1
1
1
2
208.23 g/mol
142.05 g/mol
233.40 g/mol
4.36 g
2.62 g BaSO
2.62 g
Amount of BaSO4 = 233.40 g/mol4 = 0.0112 mol
Amount of BaCl2 required to produce 0.0112 mol BaSO4 = 0.0112 1mol × 1
= 0.0112 mol
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MHR • Unit 2 Chemical Quantities
Mass of BaCl2 expected to have produced 2.62 g BaSO4
= 0.0112 mol BaCl2 × 208.23 g/mol = 2.34 g
2.34 g
Percentage purity of the original barium chloride = 4.36 g × 100% = 53.7%
17. (a)
C6H6( )
+Br2( )
C6H5 Br( )
+HBr(g)
1
1
1
1
mole ratio
molar mass
78.12 g/mol
157.01 g/mol
7.50 g
m
given/required
Amount of C6H6 =
7.50 g C6H6
78.12 g/mol
= 0.0960 mol
×1
Amount of C6H5Br produced = 0.096 mol
= 0.0960 mol
1
m = 0.0960 mol C6H5Br × 157.01 g/mol = 15.1 g C6H5Br
Therefore, the maximum amount of C6H5Br that can be formed is 15.1 g.
(b)
C6H6( )
mole ratio
1
molar mass
78.12 g/mol
C6H4Br2( )
+2Br2( )
1
2
+2HBr(g)
2
235.90 g/mol
1.25 g
given
Amount of C6H4Br2 formed in the competing reaction
1.25 g C H Br
6 4 2
= 0.00530 mol
= 235.90 g/mol
Amount of C6H6 participated in the competing reaction
= 0.005301mol × 1 = 0.00530 mol
Mass of C6H6 converted to C6H4Br2 in the competing reaction
= 0.00530 mol C6H6 × 78.12 g/mol = 0.414 g C6H6
Therefore, the mass of C6H6 not converted to C6H5Br is 0.414 g.
(c) Amount of C6H6 used to produce C6H5Br = 0.0960 mol − 0.00530 mol
= 0.0907 mol
Amount of C6H5Br produced = 0.0907 1mol × 1 = 0.0907 mol
Mass of C6H5Br produced = 0.0907 mol × 157.01 g/mol = 14.2 g
14.2 g
× 100% = 94.0%
(d) Percentage yield of C6H5Br =
15.1 g
18. In the experiment, the thermal decomposition of malachite should be used to obtain
copper(II) oxide. To determine a mole-to-mole ratio, the initial mass of malachite
should be determined and converted to moles. The CuO should be collected and its
massed measured upon completion of the reaction. The number of moles of CuO
can be calculated from the mass data and a molar ratio can be generated. This mole
ratio should be converted to integer values to make it of use in comparisons to the
balanced chemical equation for the decomposition of malachite. Safety issues for the
lab should include: the removal of CO2 from the lab, heat resistant equipment (high
temperatures, 200°C) and the handling of the CuO after the reaction has been
completed.
19. (a) Amount of A = 0.25 mol
A → B expect 0.25 × 0.60 = 0.15 mol of B
B → C expect 0.15 × 0.60 = 0.090 mol of C
C → D expect 0.09 × 0.60 = 0.054 mol of D
D → E expect 0.054 × 0.60 = 0.032 mol of E
Theoretical yield of E = 0.032 mol E × 100 g/mol = 3.2 g E
(b) From (a) 0.032 mol of E is produced from 0.25 mol of A.
× 0.25 mol
= 5.5 mol A
0.70 mol of E will be produced from 0.70 mol
0.032 mol
Chapter 7 Quantities in Chemical Reactions • MHR
283
UNIT 2 Chapter 7 QUANTITIES IN CHEMICAL REACTIONS
Mass of A required = 5.5 mol A × 200 g/mol = 1100 g A
Answers to Communication Questions
20. A hand-knit sweater requires nine balls of blue yarn, two balls of white yarn, and one
ball of red yarn. Suppose you have 50 balls of blue yarn, 15 balls of white yarn, and
three balls of red yarn. You can only knit three sweaters because each sweater requires
one ball of red yarn. The amount of red yarn you have limits the amount of sweaters
you can knit. The other two colours are in excess.
21. 2A + B → 3C + D
Determine which reactant is limiting:
Mass A (g)
n mol A = Molar mass A (g/mol)
NB
n mol B = Avogadro’s
constant
n mol C produced based on:
A → n = 32 × n mol A
B → n = 31 × n mol B
Assume that A is the limiting reagent (i.e., n mol C produced based on A < n mol C
produced based on B).
Determine the mass of C produced:
Mass A (g)
Molar mass A (g/mol)
Amount of C (mol)
C
= 32 mol
n mol A
mol A
n mol A =
Mass of C (g) = amount of C (mol) × molar mass of C (g/mol)
22. Students should include ideas and equations dealt with in Chapter 6 to complete this
written project.
Answers to Making Connections Questions
23.
2C8H18( )
+252(g)
16CO2(g)
+18H2O(g)
mole ratio
2
25
16
18
molar mass
114.26 g/mol
32.00 g/mol
given/required
m
10 L
= 67 L
Volume of octane needed to make the trip = 670 km × 100
km
Mass of octane needed = 67000 mL × 0.703 g/mL = 47 000 g
47000 g
Amount of octane needed = 114.26 g/mol = 412 mol
× 25
= 5150 mol O2
Amount of O2 consumed = 412 mol
2
Mass of O2 required = 5150 mol O2 × 32.0 g/mol = 16500 g O2 , or
165 kg of O2
To find the mass of air required, two more assumptions need to be made. (See answer
to the Mind Stretch, on page 264 of the student textbook.)
1 mol O2 occupies 24 L
5150 mol O2 occupies 1.2 × 105 L
5
Volume of air required = 1.2 × 104 L × 5 = 1.5 × 105 L
Mass of air required = 1.5 × 105 L × 1.21 g/L = 1.8 × 105 g , or 180 g of air.
24. (a) The reaction of mercury(II) nitrate with sodium sulfide can be used to remove
mercury ions from the waste water because of the formation of solid mercury(II)
284
MHR • Unit 2 Chemical Quantities
(b)
(c)
(d)
(e)
sulfide. The solid precipitate can then be filtered to separate the mercury from the
waste water.
Mercury(II) nitrate is more of an environmental concern because it is soluble in
water and can be ingested by marine species and carried up through the food
chain more easily than solid mercury(II) sulfide.
In this reaction, we assume that sodium sulfide and sodium nitrate are non-toxic
and, therefore, safe to use in the removal of toxic mercury.
If every litre of water contains 0.03 g of Hg(NO3)2 , then 10 000 L of waste water
contains 300 g Hg(NO3)2 .
Amount of Hg(NO3)2 in 10 000 L of waste water
300 g
= 324.61 g/mol = 0.924 mol Hg(NO3)2
mol Na2S consumed in the reaction Na2S and Hg(NO3)2 react in a
1 : 1 ratio)= 0.924 mol
Mass Na2S required to remove 300 g mercury ions
= 0.924 mol × 78.05 g/mol = 72.1 g = 0.0721 kg
The method should be economically feasible. The reactants chosen for the
method and the products they form should be environmentally safe and readily
available. The method should be efficient at removing all of the mercury ions. The
method chosen should still be effective in large scales.
Chapter 7 Quantities in Chemical Reactions • MHR
285