Download 5.4 CALCULATING MASSES OF REACTANTS AND PRODUCTS

→ 187
76Os +
(c)
187 Re
75
(d)
11 B+ 1 p
5
1
(e)
98
42 Mo
0
–1e
→ 3 42He
1
+ 21H → 99
43Tc + 0n
7. The mass number drops by 28, and only alpha decay drops the mass number, and each alpha particle is 4, so 7 alpha
particles are emitted. The atomic number drops by 10, which is 4 less than the drop of 14 caused by the alpha decays,
so there must be 4 beta particles emitted, because each one raises the atomic number by 1.
Making Connections
8. A typical report would include information on directing a particle beam through the body to cause maximum damage
at a tumour location — and probably on the development of the “cobalt bomb” cobalt-60 cancer therapy machines by
AECL, and the extensive worldwide use of this treatment program.
GO TO www.science.nelson.com, Chemistry 11, Teacher Centre.
5.4 CALCULATING MASSES OF REACTANTS AND PRODUCTS
PRACTICE
(Page 227)
Understanding Concepts
1. 2 Al2O3(s)
→
3 O2(g) 125 g
101.96 g/mol
1 mol
nAl O
= 125 g 2 3
101.96 g
nAl O
= 1.23 mol
2 3
4
nAl
= 1.23 mol 2
nAl
= 2.45 mol
26.98 g
mAl
= 2.45 mol 1
mol
mAl
= 66.2 g
or
mAl
mAl
4 Al(s)
m
26.98 g/mol
1
mol A
l2O3
l
26.98 g Al
4
mol A
= 125 g Al
2O3 l2O3
1
mol A
101.96 g A
l2O3 2 mol A
l
= 66.2 g
The (maximum) mass of aluminum that can be produced is 66.2 g.
2. C3H8(g) 5 O2(g) → 3 CO2(g) 4 H2O(g)
10.0 g
m
44.11 g/mol 32.00 g/mol
1 mol
nC H
= 10.0 g 3 8
44.11 g
nC H
= 0.227 mol
3 8
5
nO
= 0.227 mol 2
1
nO
= 1.13 mol
2
32.00 g
mO
= 1.13 mol 2
1
mol
120
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
mO
= 36.3 g
mO
1
mol C
5
mol O2
32.00 g O
3H8
= 10.0 g C
2
3H8 44.11 g C
H
1
m
o
l
C
H
1
mol O
2
3 8
3 8
= 36.3 g
2
or
2
mO
2
The (minimum) mass of oxygen required is 36.3 g.
3. 2 NaCl(aq) Pb(NO3)2(aq) → 2 NaNO3(aq) PbCl2(s)
2.57 g
m
58.44 g/mol
278.10 g/mol
1 mol
nNaCl
= 2.57 g 58.44 g
nNaCl
= 0.0440 mol
1
nPbCl
= 0.0440 mol 2
2
nPbCl
= 0.0220 mol
2
278.10 g
mPbCl
= 0.0220 mol 2
1
mol
mPbCl
= 6.11 g
2
or
1
mol P
bCl
278.10 g PbCl2
1m
ol N
aCl
mPbCl = 2.57 g NaCl
2 2
58.44 g N
aCl
2
mol N
aCl
1
mol P
bCl2
mPbCl
= 6.11 g
2
The (maximum) mass of lead(II) chloride produced would be 6.11 g.
4. 2 Al(s)
3 H2SO4(aq)
→ 3 H2(g)
2 Al2(SO4)3(aq)
2.73 g
m
26.98 g/mol
2.02 g/mol
1 mol
nAl
= 2.73 g 26.98 g
nAl
= 0.101 mol
3
nH
= 0.101 mol 2
2
nH
= 0.152 mol
2
2.02 g
mH
= 0.152 mol 2
1
mol
mH
= 0.307 g
2
or
mH
2
mH
2
3
mol H
2.02 g H
l
1
mol A
= 2.73 g Al
2 2
2
mol A
1
mol 26.98 g A
l
l
H2
= 0.307 g
The (predicted) mass of hydrogen would be 0.307 g, or 307 mg.
5. 2 KOH(aq) Cu(NO3)2(aq)
→ 2 KNO3(aq)
Cu(OH)2(s)
2.67 g
m
56.11 g/mol
97.57 g/mol
1 mol
nKOH
= 2.67 g 56.11 g
nKOH
= 0.0476 mol
1
nCu(OH) = 0.0476 mol 2
2
nCu(OH) = 0.0238 mol
2
97.57 g
mCu(OH) = 0.0238 mol 2
1
mol
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
121
mCu(OH) = 2.32 g
2
or
1
mol Cu(
OH)
1m
ol K
OH
mCu(OH) = 2.67 g KOH
2
2
56.11 g K
OH
2
mol K
OH
mCu(OH) = 2.32 g
97. 57 g Cu(OH)
2
1
mol C
u(OH)2
2
The mass of copper(II) hydroxide produced would be 2.32 g.
6. 2 H2O(l) → 2 H2(g)
O2(g)
25.0 g
N
18.02 g/mol
6.02 × 1023 molecules/mol
1 mol
nH O
= 25.0 g 2
18.02 g
nH O
= 1.39 mol
2
1
nO
= 1.39 mol 2
2
nO
= 0.694 mol
6.02 × 1023 molecules
1 mol
= 4.18 × 1023 molecules
2
= 0.694 mol NO
2
NO
2
or
NO
6.02 × 1023 molecules O2
1
mol H
1m
ol O2
2O
= 25.0 g H2O 18.02 g H2O
2
mol H2O
1 mol
O2
NO
= 4.18 × 1023 molecules
2
2
In this reaction, 4.18 × 1023 molecules of oxygen would be produced.
PRACTICE
(Page 228)
Understanding Concepts
7. The fundamental test for a scientific concept is its ability to predict.
8. The reaction equation coefficients show the ratio of substances in moles, so measurements must be changed to
amounts before the ratio can be applied.
9. (a) The mole ratio CO2 : C8H18 (from the equation) is 16 : 2.
(b) 2 C8H18(g) 25 O2(g)
→
16 CO2(g) 18 H2O(g)
22.8 g
m
114.26 g/mol
44.01 g/mol
1 mol
nC H
= 22.8 g 8 18
114.26 g
nC H
= 0.200 mol
8 18
16
nCO
= 0.200 mol 2
2
nCO
= 1.60 mol
2
44.01 g
mCO
= 1.60 mol 2
1
mol
mCO
= 70.3 g
2
or
mCO
2
mCO
2
1
mol C
16 mol C
O2
44.01 g CO2
8H18
= 22.8 g C
8H18 114.26 g C
2
mol C
1m
ol C
O2
8H18
8H18
= 70.3 g
The mass of carbon dioxide produced will be 70.3 g.
(c) The mole ratio O2 : C8H18 (from the equation) is 25 : 2.
122
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(d) 2 C8H18(g)
22.8 g
114.26 g/mol
nC H
8 18
nC H
8 18
nO
2
nO
2
mO
2
mO
2
or
mO
2
mO
2
25 O2(g)
→
16 CO2(g)
m
32.00 g/mol
1 mol
= 22.8 g 114.26 g
= 0.200 mol
25
= 0.200 mol 2
= 2.49 mol
32.00 g
= 2.49 mol 1
mol
= 79.8 g
18 H2O(g)
1
mol C
25 mol O2
32.00 g O
8H18
= 22.8 g C
2
8H18 114.26 g C
2
mol C
1
mol O
2
8H18
8H18
= 79.8 g
The mass of oxygen consumed will be 79.8 g.
(e) If oxygen is the limiting reagent, some carbon will not react completely to CO2(g), but instead will react incompletely to form CO(g), which is very toxic.
10. TiO2(s) 2 Cl2(g) 2 C(s) → TiCl4(l) 2 CO(g)
1.00 kg
79.88 g/mol
TiCl4(l)
2 Mg(s)
→
Ti(s) 2 MgCl2(s)
m
47.88 g/mol
Note: This question requires the students to see that the mole ratio TiCl4 : TiO2 from the first equation can be
combined with the mole ratio Ti : TiCl4 from the second equation.
nTiO
2
nTiO
2
nTiCl
4
nTiCl
4
nTi
nTi
mTi
mTi
or
1 mol
= 1.00 kg 79.88 g
= 0.0125 kmol
1
= 0.125 kmol (mole ratio TiCl4 : TiO2 is 1 : 1 — step 1)
1
= 0.125 kmol
1
= 0.125 kmol (mole ratio Ti : TiCl4 is 1 : 1 — step 2)
1
= 0.125 kmol
47.88 g
= 0.125 kmol
1
mol
= 0.599 kg or 599 g
mTi
1
mol T
iO 2
1
mol T
i Cl
i
47.88 g Ti
1
mol T
= 1.00 kg TiO2 4 1
mol T
79.88 g T
iO2
1
mol T
iO2
1
mol T
iC l4
i
mTi
= 0.599 kg or 599 g
The (maximum) mass of titanium produced would be 599 g.
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
123
SECTION 5.4 QUESTIONS
(Page 229)
Understanding Concepts
1. Reaction equations read in amounts, not masses, so combining reagents by mass according to equation values will not
work.
2. (a) Fe2O3(s) 3 H2(g)
→
2 Fe(s) 3 H2O(g)
500 kg
m
159.70 g/mol
55.85 g/mol
1 mol
nFe O
= 500 kg 2 3
159.70 g
nFe O
= 3.13 kmol
2 3
2
nFe
= 3.13 kmol 1
nFe
= 6.26 kmol
55.85 g
mFe
= 6.26 kmol
1
mol
mFe
= 350 kg
or
mFe
mFe
1
mol Fe
2
mol Fe
55.85 g Fe
2O3
= 500 kg Fe
2O3 159.70 g F
e2O3 1 mol F
e2O3
1
mol F
e
= 350 kg
The mass of iron produced would be 350 kg.
(b) Fe2O3(s)
3 H2(g) →
2 Fe(s) 1.0 t = 1.0 Mg
m
159.70 g/mol
2.02 g/mol
1 mol
nFe O
= 1.0 Mg 2 3
159.70 g
nFe O
= 0.0063 Mmol = 6.3 kmol
2 3
3
nH
= 0.0063 Mmol 2
1
nH
= 0.019 Mmol
2
2.02 g
mH
= 0.019 Mmol
2
1
mol
mH
= 0.038 Mg = 38 kg
3 H2O(g)
2
or
mH
2
mH
2
1
mol Fe
3m
ol H2
2.02 g H
2O3
= 1.0 Mg Fe
2
2O3 159.70 g F
e2O3 1 mol F
e2O3 1 mol H2
= 0.038 Mg = 38 kg
The mass of hydrogen required would be 38 kg.
(c) Fe2O3(s) 3 H2(g)
→
2 Fe(s)
m
159.70 g/mol
1 mol
nH O
= 220 kg 2
18.02 g
nH O
= 12.2 kmol
2
1
nFe O
= 12.2 kmol 2 3
3
nFe O
= 4.07 kmol
2 3
159.70 g
mFe O = 4.07 kmol
2 3
1
mol
124
Unit 2 Quantities in Chemical Reactions
3 H2O(g)
220 kg
18.02 g/mol
Copyright © 2002 Nelson Thomson Learning
mFe O
= 650 kg
mFe O
1
mol H
1
mol F
e2O3 159.70 g Fe2O3
2O
= 220 kg H2O 18.02 g H2O
3m
ol H2O
1
mol F
e2O3
= 650 kg
2 3
or
2 3
mFe O
2 3
The mass of iron(III) oxide consumed would be 650 kg.
3. (a) 2 C6H12(l) 5 O2(g)
→
2 C6H10O4(s) m
280 kg
84.18 g/mol
146.16 g/mol
1 mol
nC H O = 280 kg 6 10 4
146.16 g
nC H O = 1.92 kmol
6 10 4
2
nC H
= 1.92 kmol 6 12
2
nC H
= 1.92 kmol
6 12
84.18 g
mC H
= 1.92 kmol
6 12
1
mol
mC H
= 161 kg
2 H2O(g)
6 12
or
mC H
6 12
mC H
6 12
1
mol C6
H10O4
2m
ol C
84.18 g C6H12
6H12
= 280 kg C6
H10O4 146.16 g C6
H10O4
2
mol C6
H10O4
1
mol C
6H12
= 161 kg
The mass of cyclohexane required is 161 kg.
(b) 2 C6H12(l) 5 O2(g)
→
2 C6H10O4(s)
125 kg
m
84.18 g/mol
32.00 g/mol
1 mol
nC H
= 125 kg 6 12
84.18 g
nC H
= 1.48 kmol
6 12
5
nO
= 1.48 kmol 2
2
nO
= 3.71 kmol
2
32.00 g
mO
= 3.71 kmol
2
1
mol
mO
= 119 kg
2 H2O(g)
2
or
mO
2
mO
2
1m
ol C
5
mol O2
32.00 g O
6H12
= 125 kg C
2
6H12 84.18 g C
H
2
m
ol
C
H
1
mol O
2
6 12
6 12
= 119 kg
The mass of oxygen required is 119 kg.
(c) 2 C6H12(l) 5 O2(g)
→
2 C6H10O4(s)
m
200 g
84.18 g/mol
32.00 g/mol
1 mol
nO
= 200 g 2
32.00 g
nO
= 6.25 mol
2
2
nC H
= 6.25 mol 6 12
5
nC H
= 2.50 mol
2 H2O(g)
6 12
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
125
mC H
6 12
mC H
6 12
or
mC H
6 12
mC H
6 12
84.18 g
= 2.50 mol 1
mol
= 210 g
1
mol O
2
2
mol C
84.18 g C6H12
6H12
= 200 g O2 32.00 g O2
5
mol O2
1
mol C
6H12
= 210 g
The maximum mass of cyclohexane that can be reacted is 210 g.
4. 2 Fe(s)
O2(g)
→
2 FeO(s)
56.8 g
m
55.85 g/mol
71.85 g/mol
1 mol
nFe
= 56.8 g 55.85 g
nFe
= 1.02 mol
2
nFeO
= 1.02 mol 2
nFeO
= 1.02 mol
71.85 g
mFeO
= 1.02 mol 1
mol
mFeO
= 73.1 g
or
mFeO
mFeO
e
2
mol FeO
71.85 g FeO
1
mol F
= 56.8 g Fe
2
mol Fe
1
mol F
eO
55.85 g Fe
= 73.1 g
The mass of iron(II) oxide that can be produced is 73.1 g.
Applying Inquiry Skills
5. (a) Prediction
From the balanced reaction equation, one mole of calcium ions produces one mole of solid calcium oxalate
precipitate, so the amount of calcium ions in a sample will be the same as the amount of calcium oxalate precipitate that forms.
(b) Experimental Design
Excess sodium oxalate solution will be added to a measured sample (independent variable) of hard water. The
precipitate formed will be dried and the mass measured (dependent variable). The stoichiometric method will be
used to calculate an answer to the question.
(c) The mass of calcium oxalate precipitate will be divided by its molar mass, 128.10 g/mol, to determine the amount
of precipitate. The amount of calcium ions is the same, since the mole ratio is 1 : 1.
(d) Materials
•
hard-water sample
126
•
sodium oxalate solution
•
wash bottle with pure water
•
two 100-mL graduated cylinders
•
250-mL beaker
•
filtration apparatus
•
filter paper
•
centigram balance
Unit 2 Quantities in Chemical Reactions
Copyright © 2002 Nelson Thomson Learning
(e) Procedure
1. Use a graduated cylinder to add 10.0 mL of hard water to a 250-mL beaker.
2. Use another graduated cylinder to add about 10 mL of sodium oxalate solution to the beaker.
3. Allow the precipitate to settle.
4. Add another 10-mL aliquot of sodium oxalate solution to the beaker, observing whether or not more precipitate forms as the added solution mixes with the clear upper portion of the solution in the beaker.
5. Repeat step 4 until no more precipitate forms.
6. Measure and record the mass of a piece of filter paper.
7. Filter, wash, and dry the precipitate.
8. Measure and record the mass of the filter paper plus dry precipitate.
9. Dispose of all materials as instructed.
Making Connections
6. (a) Raw materials, like aluminum oxide, are purchased according to the amount needed for reaction.
(b) Products, like aluminum, are priced so that the amount produced will be profitable.
(c) Pollutant amounts will be calculated and decisions on processes to control them will be made accordingly.
(d) Amounts of all materials must be calculated to allow costing of things like packaging, disposal, and shipping.
5.5 CALCULATING LIMITING AND EXCESS REAGENTS
PRACTICE
(Page 231)
Understanding Concepts
1. (a) 2 CH4(g) 3 O2(g)
→
m
16.05 g/mol
3.0 mol
nCH
4
nCH
4
mCH
4
mCH
4
mCH
4
mCH
4
or
2 CO(g)
4 H2O(g)
2
= 3.0 mol 3
= 2.0 mol
16.05 g
= 2.0 mol 1
mol
= 32 g
2
mol CH4
16.05 g CH4
= 3 mol
O2 3
mol O2
1
mol C
H4
= 32 g
The mass of methane that will react is 32 g.
(b) CH4(g)
2 O2(g)
→
CO2(g)
m
16.05 g/mol
3.0 mol
1
nCH
= 3.0 mol 4
2
nCH
= 1.5 mol
4
16.05 g
mCH
= 1.5 mol 4
1
mol
mCH
= 24 g
2 H2O(g)
4
or
mCH
4
1
mol CH4
16.05 g CH4
= 3 mol
O2 2
mol O2
1
mol C
H4
Copyright © 2002 Nelson Thomson Learning
Chapter 5 Quantities in Chemical Equations
127