Chapter 7: Stoichiometry in Chemical Reactions
Section 7.1: Mole Ratios in Chemical, pages 113–114
1. In the equation 4 Al(s) + 3 O2(g) → 2 Al2 O3(s), there is a ratio of 3:2 between the amount of oxygen used
and the amount of aluminum(III) oxide produced.
2. False. The total mass of the reactants on the left side of a chemical equation always equals the total mass of
the products on the right side of the equation.
3. c
4. (a) Table 1 Buffalo Wing Synthesis
6B
+
4R
+
4K
+
1H
→
1 Bf
B
R
K
H
Bf
6
4
4
1
1
9
6
6
1.5
1.5
90
60
60
15
15
6 × 6.02 × 1023 4 × 6.02 × 1023
4 × 6.02 × 1023
1 × 6.02 × 1023
1 × 6.02 × 1023
= 6 mol
= 4 mol
= 4 mol
= 1 mol
= 1 mol
(b) The number of entities on both sides of the equation are not equal because the product, Bf, has more mass
per entity than the reactants.
(c) The total mass of the reactants in the equation equals the total mass of the product.
5. Given: nH O = 3.25 mol
2
Required: amount of HCl, nHCl
Solution:
Step 1: Write a balanced equation for the reaction.
MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O
nHCl
3.25 mol
Step 2: Convert amount of H2O to amount of HCl.
nHCl = 3.25 mol H 2O !
4 mol HCl
2 mol H 2O
nHCl = 6.50 mol
Statement: To produce 3.25 mol of water, 6.50 mol of HCl will be needed.
6. Given: nNO = 2.5 " 10!1 mol
Required: amount of HNO2, nHNO
2
Solution:
Step 1: Write a balanced equation for the reaction.
3 HNO2 → HNO3 + H2O + 2 NO
nHNO
2.5 × 10–1 mol
2
Step 2: Convert amount of NO to amount of HNO2.
nHNO2 = 2.5 " 10!1 mol NO "
3 mol HNO2
2 mol NO
nHNO2 = 3.8 " 10 mol
!1
Statement: To produce 2.5 × 10–1 mol of NO, 3.8 × 10–1 mol of HNO2 will be needed.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-1
Section 7.2: Mass Relationships in Chemical Equations, pages 115–116
1. The amount of a substance in moles, n, can be determined by dividing the mass of the substance by its molar
mass.
2. In the equation CO2(g) + 2 LiOH(s) → Li2CO3(aq) + 2 H2O(l), the stoichiometric relationship of lithium
hydroxide to carbon dioxide tells us that 2 moles of LiOH are needed to react with 1 mole of CO2.
3. True
4. Answers may vary. Sample answer:
Figure 1
5. Given: mO = 148.0 g
2
Required: mass of NH3, mNH
3
Solution:
Step 1: Write a balanced equation for the reaction.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
mNH
148.0 g
3
Step 2: Convert mass of O2 to amount of O2.
nO2 = 148.0 g !
1 mol O2
32.00 g
nO2 = 4.625 mol
Step 3: Convert amount of O2 to amount of NH3.
nNH3 = 4.625 mol O2 !
4 mol NH3
5 mol O2
nNH3 = 3.700 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-2
Step 4: Convert amount of NH3 to mass of NH3.
! 17.04 g "
mNH3 = (3.700 mol ) #
$
% 1 mol &
mNH3 = 63.05 g
Statement: To react with 148.0 g of oxygen, 63.05 g of ammonia is required.
6. Given: mIF = 400.0 g
3
Required: mass of F2, mF
2
Solution:
Step 1: Write a balanced equation for the reaction.
I2(s) + 3 F2(g) → 2 IF3(s)
Step 2: Convert mass of IF3 to amount of IF3.
nIF = 400.0 g !
3
1 molIF
3
183.90 g
nIF = 2.175 10 mol (two extra digits carried)
3
Step 3: Convert amount of IF3 to amount of F2.
nF = 2.175 10 molIF !
2
3
3 molF
2
2 molIF
3
nF = 3.262 65 mol
2
Step 4: Convert amount of F2 to mass of F2.
! 38.00 g $
mF = (3.262 65 mol ) #
&
2
" 1 mol %
mF = 124.0 g
2
Statement: 124.0 g of fluorine gas is required to produce 400.0 g of iodine(III) fluoride.
Section 7.3: Which Reagent Runs Out First?, pages 117–118
1. False. The compound or element that is greater in terms of stoichiometric amount is the excess reagent in a
reaction.
2. d
3. (a) The process of making the salad dressing (S) from olive oil (O) and vinegar (V) is described by the
equation 2 O + 1 V → 1 S
Since the ratio of olive oil to vinegar is 2:1, only 50 mL of vinegar is required to combine with 100 mL of
olive oil. Therefore, the limiting reagent (ingredient) is olive oil. The excess reagent is vinegar.
(b) The maximum amount of properly proportioned salad dressing that the chef can make is
100 mL + 50 mL = 150 mL
(c) 10 mL of vinegar will be left over.
(d) To make the limiting reagent (olive oil) to become the excess reagent, a stoichiometric amount of 20 mL of
olive oil is required to use up the 10 mL of vinegar that is left over. To make olive oil the excess reagent, you
would need to add more than 20 mL of olive oil.
4. a
5. Table 1 Combustion of Gasoline: C7 H16(g) + 11 O2(g) → 7 CO2(g) + 8 H2O(g)
Amount of
Amount of O2
Mole ratio
Complete
Excess reagent
C7H16 (mol)
(mol)
combustion?
nO : nC H
2
5
7
3 × 10–2
2.3 × 103
5
56
5 × 10–1
9.2 × 102
Copyright © 2011 Nelson Education Ltd.
7
1:1
8:1
17 : 1
0.4 : 1
16
no
no
yes
no
C7H16
C7H16
O2
C7H16
Chapter 7: Stoichiometry in Chemical Reactions
7-3
6. (a) Complete combustion produces no air pollutants. Incomplete combustion produces both carbon
monoxide and soot.
(b) Complete combustion produces 3 moles of carbon dioxide for 1 mole of propane. The mole ratio of
propane to carbon dioxide is 1:3. Incomplete combustion produces 2 moles of carbon dioxide for 2 moles of
propane. The mole ratio of propane to carbon dioxide is 1:1. So, for 1 mole of propane, complete combustion
produces more carbon dioxide (the primary greenhouse gas).
7. (a) The acid is the limiting reagent in this situation. You want to make the base the excess reagent to make
sure that all of the acid is neutralized.
(b) Stoichiometric ratios need to be considered when cleaning up an acid spill because when too much base is
added, a new problem will develop—caustic base left in the pond.
Section 7.4: Calculations Involving Limiting Reagents, pages 119–120
1. If the actual mole ratio of reactant A to reactant B in a reaction is greater than the ideal stoichiometric ratio,
then reactant A is the excess reagent.
2. If the actual mole ratio of reactant A to reactant B in a reaction is less than the ideal stoichiometric ratio,
then reactant A is the limiting reagent.
3. d
4. (a) If X is the limiting reagent, you would find Y and Z in the container after the reaction was complete.
(b) If X is the excess reagent, you would find X and Z in the container after the reaction was complete.
(c) If the reaction is carried out to precise stoichiometric values, you would find only the product Z in the
container after the reaction was complete.
5. Given: nK = 7.0 mol ; nO = 2.4 mol
2
Required: mass of tarnish, mK O
2
Solution:
Step 1: Write a balanced equation for the reaction.
4 K(s) + O2(g) → 2 K2O(s)
7.0 mol 2.4 mol mK O
2
Step 2: Determine the limiting reagent. Use the amount of one reactant to find the stoichiometric amount of
another.
nK = 2.4 mol O2 !
4 mol K
1 mol O2
nK = 9.6 mol
Since the amount of K present initially is less than the required amount, K is the limiting reagent.
Step 3: Convert amount of K to amount of K2O.
nK2O = 7.0 mol K !
2 mol K2O
4 mol K
nK2O = 3.5 mol
Step 4: Convert amount of K2O to mass of K2O.
! 94.20 g "
mK2O = (3.5 mol ) #
$
% 1 mol &
mK2O = 330 g
Statement: When 7.0 moles of K reacts with 2.4 moles of O2, 300 grams of K2 O will form.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-4
6. (a) Reagent A and the product A2B3 will be left in the container when the reaction is complete.
(b) Reagent B is the limiting reagent. Convert amount of B to amount of product A2B3.
nA2 B3 = 6.0 mol B !
1 mol A2 B3
3 mol B
nA2 B3 = 2.0 mol
An amount of 2.0 moles of the product A2B3 would be produced from the conditions in part (a).
7. Given: mC = 20.2 g ; mK CO = 52.5 g
2
3
Required: mass of potassium, mK
Solution:
Step 1: Write a balanced equation for the reaction.
C(s)
+
2 K2CO3(s) → 4 K(s) + 3 CO2(g)
mK
20.2 g
52.5 g
12.01 g/mol 138.21 g/mol
Step 2: Convert mass of given substances to amount.
nC = 20.2 g !
1 mol C
12.01 g
nC = 1.6819 mol (two extra digits carried)
nK CO = 52.5 g !
2
3
1 molK CO
2
3
138.21 g
nK CO = 0.379 86 mol (two extra digits carried)
2
3
Step 3: Determine the limiting reagent. Use the amount of one reactant to find the stoichiometric amount of
another.
nK2CO3 = 1.6819 mol C !
2 mol K2CO3
1 mol C
nK2CO3 = 3.3638 mol (two extra digits carried)
Since the amount of K2CO3 present initially is less than the required amount, K2CO3 is the limiting reagent.
Step 4: Convert amount of K2CO3 to amount of K.
nK = 0.379 68 molK CO !
2
3
4 molK
2 molK CO
2
3
nK = 0.759 36 mol (two extra digits carried)
Step 5: Convert amount of K to mass of K.
! 39.10 g $
mK = (0.759 36 mol ) #
&
" 1 mol %
mK = 29.7 g
Statement: When 20.2 g of carbon reacts with 52.5 g of K2CO3, 29.7 g of potassium is produced.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-5
Section 7.5: Percentage Yield, pages 121–123
1. The actual yield of a reaction is usually lower than the theoretical yield.
2. True
actual yield
3. (a) percentage yield =
theoretical yield
32 g
=
! 100 %
40 g
! 100 %
percentage yield = 80 %
If the reaction yielded 32 g of the product, the percentage yield was 80 %.
(b) percentage yield =
actual yield
theoretical yield
42 g
=
! 100 %
40 g
! 100 %
percentage yield = 105 %
If the reaction yielded 42 g of the product, the percentage yield was 105 %.
(c) The impurities of the reactant might react to produce the same product. This would cause a yield that was
greater than the theoretical yield in part (b), giving the percentage yield a value of over 100 %.
4. Table 1 Why Actual Yield Can Be Reduced
Reason for lower
actual yield
What happens
Example
Result
reversible reactions
a reaction product starts
reacting on its own
spills, evaporation,
small quantities of reactants
splatter, loss through
are lost before they can
transfer
react
impurities
some of the reactant is the
wrong chemical
completing side
a reaction occurs that yields
reactions
the wrong product
5. Given: nAl = 3.8 mol ; nAlCl3 = 1.6 mol
ammonia decomposes after
it is formed
reactant is left in a container
during a transfer
lower yield of ammonia than
expected
lower yield of product than
expected
sodium hydroxide absorbs
water
transplantin is created
instead of cisplantin
lower reactivity of impure
NaOH
lower yield of cisplantin than
expected
Required: percentage yield of AlCl3
Solution:
Step 1: Write a balanced equation for the reaction.
2 Al(s) + 3 Cl2 (g) → 2 AlCl3 (s)
3.8 mol
1.6 mol
Since the mole ratio of Al to AlCl3 is 2:2, or 1:1, the theoretical yield of AlCl3 is 3.8 mol.
Step 2: Calculate the percentage yield.
percentage yield =
=
actual yield
theoretical yield
1.6 mol
3.8 mol
percentage yield = 42 %
! 100 %
! 100 %
Statement: If 3.8 mol of aluminum reacts with an excess of Cl2 to produce 1.6 mol of AlCl3, the percentage
yield of the reaction is 42 %.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-6
6. Given: nFe = 234.0 g ; actual yield of Fe2O3 = 310.6 g
Required: percentage yield of Fe2 O3
Solution:
Step 1: Write a balanced equation for the reaction.
4 Fe(s) + 3 O2(g) → 2 Fe2 O3(s)
mFe O
234.0 g
2
3
55.85 g/mol
159.70 g/mol
Step 2: Convert mass of Fe to amount of Fe.
nFe = 234.0 g !
1 molFe
55.85 g
nFe = 4.189 79 mol (two extra digits carried)
Step 3: Convert amount of Fe to amount of Fe2O3.
nFe O = 4.189 79 molFe !
2
2 molFe O
3
2
3
4 molFe
nFe O = 2.094 90 mol (two extra digits carried)
2
3
Step 4: Convert amount of Fe2O3 to mass of Fe2O3.
! 159.70 g $
mFe O = (2.094 90 mol ) #
&
2 3
" 1 mol %
mFe O = 334.556 g (two extra digits carried)
2
3
Step 5: Calculate the percentage yield of Fe2O3.
actual yield
! 100 %
theoretical yield
310.6 g
=
! 100 %
334.556 g
percentage yield =
percentage yield = 92.84 %
Statement: If 234.0 g of iron reacts with an excess of oxygen to produce 310.6 g of Fe2O3, the percentage
yield of the reaction is 92.84 %.
7. Given: mFe = 380.0 g ; mO = 111.1 g ; actual yield of Fe3O4 = 302.5 g
2
Required: percentage yield of Fe3 O4
Solution:
Step 1: Write a balanced equation for the reaction.
3 Fe(s) + 2 O2(g) → Fe3O4(s)
380.0 g
111.0 g
302.5 g
55.85 g/mol 32.00 g/mol 231.55 g/mol
Step 2: Convert mass of Fe to amount of Fe and mass of O2 to amount of O2.
nFe = 380.0 g !
1 mol Fe
55.85 g
nFe = 6.80394 mol (two extra digits carried)
nO = 111.1 g !
2
1 molO
2
32.00 g
nO = 3.471 88 mol (two extra digits carried)
2
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-7
Step 3: Determine the limiting reagent. Use the amount of one reactant to find the stoichiometric amount of
another.
nFe = 3.471 88 molO !
2
3 molFe
2 molO
2
nFe = 5.207 82 mol (two extra digits carried)
Since the amount of Fe present initially is greater than the required amount, O2 is the limiting reagent.
Step 4: Convert amount of O2 to amount of Fe3 O4.
nFe O = 3.471 88 molO !
3
4
1 molFe O
2
3
2 molO
4
2
nFe O = 1.735 94 mol (two extra digits carried)
3
4
Step 5: Convert amount of Fe3O4 to mass of Fe3O4.
! 231.55 g $
mFe O = (1.735 94 mol ) #
&
3 4
" 1 mol %
mFe O = 401.957 g (two extra digits carried)
3
4
Step 6: Calculate the percentage yield of Fe3O4.
actual yield
! 100 %
theoretical yield
302.5 g
=
! 100 %
401.957 g
percentage yield =
percentage yield = 75.26 %
Statement: If 380.0 g of iron reacts with 111.1 g of O2, to produce 302.5 g of Fe3O4, the percentage yield of
the reaction is 75.26 %.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-8
Chapter 7 Summary, page 124
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-9
Chapter 7 Questions, pages 125–126
1. (a) P4(s) + 5 O2(g) → 2 P2O5(g)
(b) The mole ratio between O2 and P2O5 in the equation for part (a) is 5:2.
(c) You would need 0.5 moles of P4 to produce 1 mole of P2O5.
2. (a) False. If you use stoichiometric amounts in a reaction, you will have no reactants left when the reaction
is complete.
(b) True
3. b
4. Given: mCO = 100.0 g
2
Required: mass of C3 H8, mC H
3
8
Solution:
Step 1: Write a balanced equation for the reaction.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
mC H
100.0 g
3
8
Step 2: Convert mass of O2 to amount of O2.
nCO = 100.0 g !
2
1 molO
2
44.01 g
nCO = 2.272 21 mol (two extra digits carried)
2
Step 3: Convert amount of O2 to amount of C3H8.
nC H = 2.272 21 molCO !
3
8
2
1 molC H
3
3 molCO
8
2
nC H = 0.757 403 mol (two extra digits carried)
3
8
Step 4: Convert amount of C3H8 to mass of C3H8.
! 44.11 g $
mC H = (0.757 403 mol ) #
&
3 8
" 1 mol %
mC H = 33.41 g
3
8
Statement: For the combustion of propane, if 100.0 g of carbon dioxide was produced, 33.41 g of propane was
burned.
5. (a) 2 X2 + 3 Y → X4Y3
(b) Answers may vary. Sample answer: A diagram that shows a stoichiometric representation of the reaction is
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-10
(c) Answers may vary. Sample answer: A diagram that shows a reaction in which X is the limiting reagent and
Y is the excess reagent is
(d) Answers may vary. Sample answer: A diagram that shows a reaction in which Y is the limiting reagent and
X is the excess reagent is
6. (a) Given: mB = 25.7 g ; mCl = 200.0 g
2
Required: limiting reagent of reaction
Solution:
Step 1: List masses and molar masses of given substances under the balanced equation for the reaction.
2 B(s)
+ 3 Cl2 (g) → 2 BCl3 (g)
25.7 g
200.0 g
10.81 g/mol 70.90 g/mol 117.15 g/mol
Step 2: Convert mass of B to amount of B and mass of Cl2 to amount of Cl2.
nB = 25.7 g !
1 mol B
10.81 g
nB = 2.3774 mol (two extra digits carried)
nCl = 200.0 g !
2
1 molO
2
70.90 g
nCl = 2.820 87 mol (two extra digits carried)
2
Step 3: Use the amount of one reactant to find the stoichiometric amount of another.
nB = 2.82087 mol Cl2 !
2 mol B
3 mol Cl2
nB = 1.88058 mol (two extra digits carried)
Statement: Since the amount of boron present is greater than the required amount, chlorine is the limiting
reagent.
(b) Given: nB = 1.88058 mol
Required: theoretical yield of BCl3
Solution:
Step 1: Convert amount of B to amount of BCl3.
From the balanced equation, he amount of BCl3 is the same as the amount of B.
mBCl3 = 1.88058 mol (two extra digits carried)
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-11
Step 2: Convert amount of BCl3 to mass of BCl3.
! 117.15 g $
mBCl = (1.880 58 mol ) #
&
3
" 1 mol %
mBCl = 220.310 g (two extra digits carried)
3
Statement: The theoretical yield of the reaction is 220.3 g of BCl3.
(c) Given: theoretical yield = 220.310 g; actual yield = 176.4 g
Required: percentage yield of reaction
Solution:
actual yield
! 100 %
theoretical yield
176.4 g
=
! 100 %
200.310 g
percentage yield =
percentage yield = 88.06 %
Statement: The percentage yield of the reaction is 80.06 %.
7. Figure 1
Unit 3 Questions, pages 127–128
1. A qualitative analysis would be needed to determine whether the two mirror images of the new flavour
compound have different flavours.
2. c
3. False. All compounds with the same chemical formula have the same molar mass.
4. (a) Table 1 Analysis of Compounds A and B with Formula CxHyOz
Compound
Mass of
Mass of
Mass of
Mass of
compound (g) carbon (g) hydrogen (g) oxygen (g)
A
20.0
12.0
1.6
6.4
A
25.0
15.0
2.0
8.0
B
20.0
8.0
1.3
10.7
B
25.0
10.0
1.6
13.4
(b) Left compound: empirical formula: CH2 O; molecular formula: C5 H10O5
Right compound: empirical formula: C5 H8O2; molecular formula: C5H8 O2
(c) Compound A is C5H8O2 and compound B is C5H10O5.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-12
5. Figure 2
6. Given: mCrF = 200.0 g
5
Required: mass of F2, mF
2
Solution:
Step 1: Write a balanced equation for the reaction.
2 Cr(s) + 5 F2 (g) → 2 CrF5 (s)
mF
200.0 g
2
Step 2: Convert mass of CrF5 to amount of CrF5.
nCrF = 200.0 g !
5
1 molCrF
5
147.00 g
nCrF = 1.360 54 mol (two extra digits carried)
5
Step 3: Convert amount of CrF5 to amount of F2.
nF = 1.360 54 molCrF !
2
5
5 molF
2
2 molCrF
5
nF = 3.401 35 mol (two extra digits carried)
2
Step 4: Convert amount of F2 to mass of F2.
! 38.00 g $
mF = (3.401 35 mol ) #
&
2
" 1 mol %
mF = 129.3 g
2
Statement: A mass of 129.3 g of fluorine is required to produce 200.0 g of CrF5.
7. For the reaction: 4 Cu(s) + O2(g) → 2 Cu2O
First convert mass of Cu to amount of Cu and mass of O2 to amount of O2.
nCu = 250.0 g !
1 molFe
63.55 g
nCu = 3.933 91 mol (two extra digits carried)
nO2 = 48.0 g !
1 mol O2
32.00 g
nO2 = 1.50 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-13
Use the amount of one reactant to find the stoichiometric amount of another.
nCu = 1.50 mol O2 !
4 mol Fe
1 mol O2
nCu = 6.00 mol
Since the amount of Cu present initially is less than the required amount, Cu is the limiting reagent and O2 is
the excess reagent.
8. Given: mMg = 50.0 g ; mN = 30.0 g
2
Required: mMg N
3
2
Solution:
Step 1: List the masses and molar masses under the balanced equation for the reaction.
3 Mg(s) + N2 (g) → Mg3N2
mMg N
50.0 g
30.0 g
3
2
24.31 g/mol 28.02 g/mol 100.95 g/mol
Step 2: Convert mass of given substances to amount.
nMg = 50.0 g !
1 mol Mg
24.31 g
nMg = 2.0568 mol (two extra digits carried)
nN 2 = 30.0 g !
1 mol N 2
28.02 g
nN 2 = 1.0707 mol (two extra digits carried)
Step 3: Determine the limiting reagent. Use the amount of one reactant to find the stoichiometric amount of
another.
nMg = 1.0707 mol N 2 !
3 mol Mg
1 mol N 2
nMg = 3.2121 mol (two extra digits carried)
Since the amount of Mg present initially is less than the required amount, Mg is the limiting reagent.
Step 4: Convert amount of Mg to amount of Mg3N2.
nMg3 N 2 = 2.0568 mol Mg !
1 mol Mg3 N 2
3 mol Mg
nMg3 N 2 = 0.6856 mol (one extra digit carried)
Step 5: Convert amount of Mg3N2 to mass of Mg3N2.
! 100.95 g "
mMg3 N 2 = (0.6856 mol ) #
$
% 1 mol &
mMg3 N 2 = 69.2 g
Statement: A mass of 69.2 g of Mg3N2 can be produced by combining 50.0 g of Mg and 30.0 g of N2 in
stoichiometric proportions.
9. (a) For the reaction: S8(l) + 4 Cl2(g) → 4 S2Cl2(l)
First convert mass of S8 to amount of S8 and mass of Cl2 to amount of Cl2.
nS8 = 140.0 g !
1 molS8
256.56 g
nS8 = 0.545681 mol (two extra digits carried)
nCl2 = 140.0 g !
1 mol Cl2
70.90 g
nCl2 = 1.97461 mol (two extra digits carried)
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-14
Use the amount of one reactant to find the stoichiometric amount of another.
nCl2 = 0.545681 molS8 !
4 mol Cl2
1 molS8
nCl2 = 2.18272 mol
Since the amount of Cl2 present initially is less than the required amount, Cl2 is the limiting reagent.
(b) From the balanced equation S8(l) + 4 Cl2(g) → 4 S2Cl2(l), the mole ratio of Cl2 to S2Cl2 is 4, 4, or 1:1.
nS Cl = nCl = 1.974 61 mol
2
2
2
Convert amount of S2Cl2 to mass of S2Cl2.
! 135.04 g "
mS2Cl2 = (1.97461 mol ) #
$
% 1 mol &
mS2Cl2 = 266.651 g (two extra digits carried)
The theoretical yield of the reaction is 266.6 g of S2Cl2.
(c) actual yield of S2Cl2 = 58.4 g
theoretical yield of S2Cl2 = 266.651 g
percentage yield =
actual yield
! 100 %
theoretical yield
58.4 g
=
! 100 %
266.651 g
percentage yield = 21.9 %
The percentage yield of the reaction is 21.9 %.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-15