Chapter 8
Solution Stoichiometry
Note to teacher: You will notice that there are two different formats for the Sample
Problems in the student textbook. Where appropriate, the Sample Problem contains the
full set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act on
Your Strategy, and Check Your Solution. Where a shorter solution is appropriate, the
Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check
Your Solution step is also included in the shorter Sample Problems.
Solutions for Practice Problems
Student Textbook page 286
1. Problem
Decide whether each of the following salts is soluble or insoluble in distilled water.
Give reasons for your answer.
(a) lead(II) chloride, PbCl2 (a white crystalline powder used in paints)
(b) zinc oxide, ZnO (a white pigment used in paints, cosmetics, and calamine lotion)
(c) silver acetate, AgCH3COO (a whitish powder that is used to help people quit
smoking because of the bitter taste it produces)
What Is Required?
Determine if the salts listed will dissolve in water.
What Is Given?
You have the names of the three salts and access to the solubility guidelines.
Plan Your Strategy
Identify the cations and anions in each salt. Refer to the solubility chart in the
student textbook and determine the guideline solubility of these ions. The solubility
of the salt will correspond with the solubility of the ion having the higher guideline
(lower number).
Act on Your Strategy
(a) Salts of chloride are soluble, and lead (II) chloride is not an exception.
(b) The solubility of oxides is not given in the chart, but students may recognize that
zinc oxide is insoluble.
(c) Most salts of acetates are soluble, but silver acetate is an exception.
Check Your Solution
Check your results against another reference. These conclusions are correct.
2. Problem
Which of the following compounds are soluble in water? Explain your reasoning for
each compound.
(a) potassium nitrate, KNO3 (used to manufacture gunpowder)
(b) lithium carbonate, Li2CO3 (used to treat people who suffer from depression)
(c) lead(II) oxide, PbO (used to make crystal glass)
What Is Required?
Determine if the salts listed will dissolve in water.
Chapter 8 Solution Stoichiometry • MHR
145
What Is Given?
You have the names of the three salts and access to the solubility guidelines.
Plan Your Strategy
Identify the cations and anions in each salt. Refer to the solubility chart in the
student textbook and determine the guideline solubility of these ions. The solubility
of the salt will correspond with the solubility of the ion having the higher guideline
(lower number).
Act on Your Strategy
(a) All salts of nitrate and potassium are soluble. Therefore, potassium nitrate is
soluble.
(b) All salts of Group 1 cations, including lithium, are soluble. Therefore, lithium car-
bonate is soluble.
(c) The solubility of oxides is not included in the solubility table. Students will likely
predict that lead(II) oxide is insoluble since other compounts of lead seem to be
insoluble.
Check Your Solution
Check the predictions against another reference. These results are correct.
3. Problem
Which of the following compounds are insoluble in water?
(a) calcium carbonate (present in marble and limestone)
(b) magnesium sulfate, MgSO4 (found in the hydrated salt, MgSO4 • 7H2O, also
known as Epsom salts; used for the relief of aching muscles and as a laxative)
(c) aluminum phosphate, AlPO4 (found in dental cements)
What Is Required?
Determine if the salts listed will dissolve in water.
What Is Given?
You have the names of the three salts and access to the solubility guidelines.
Plan Your Strategy
Identify the cations and anions in each salt. Refer to the solubility chart in the
student textbook and determine the guideline solubility of these ions. The solubility
of the salt will correspond with the solubility of the ion having the higher guideline
(lower number).
Act on Your Strategy
Name
Formula
calcium
carbonate
CaCO3
magnesium
sulfate
MgSO4
aluminum
phosphate
AlPO4
Ions
Guideline
Ion Solubility
2+
4
insoluble
CO3
2
2
insoluble
Mg2+
Ca
5
soluble
2
5
soluble
Al3+
5
soluble
3
2
insoluble
SO4
PO4
Salt Solubility
INSOLUBLE
SOLUBLE
INSOLUBLE
Check Your Solution
Check the predictions against another reference. These results are correct.
Chapter 8 Solution Stoichiometry • MHR
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Solutions for Practice Problems
Student Textbook page 290
4. Problem
Predict the result of mixing each pair of aqueous solutions. Write a balanced chemical
equation if you predict that a precipitate forms. Write “NR” if you predict that no
reaction takes place.
(a) sodium sulfide and iron(II) sulfate
(b) sodium hydroxide and barium nitrate
(c) cesium phosphate and calcium bromide
(d) sodium carbonate and sulfuric acid
(e) sodium nitrate and copper(II) sulfate
(f) ammonium iodide and silver nitrate
(g) potassium carbonate and iron(II) nitrate
(h) aluminum nitrate and sodium phosphate
(i) potassium chloride and iron(II) nitrate
(j) ammonium sulfate and barium chloride
(k) sodium sulfide and nickel(II) sulfate
(l) lead(II) nitrate and potassium bromide
What Is Required?
Predict whether or not each pair of aqueous solutions will provide ions that will combine to form an insoluble product (precipitate). Write a balanced chemical equation if
a reaction is predicted and “NR” if no reaction is predicted.
What Is Given?
You know the names of the compound in each solution.
Plan Your Strategy
Identify the ions in each compound. Exchange the cations in the two compounds
and for each new compound, look up its solubility guideline in the student textbook.
Predict whether or not either of these new compounds is insoluble. Write the
balanced equation for those examples in which a new insoluble product formed and
“NR” if both the new compounds are soluble.
Act on Your Strategy
Solubility Key: S = soluble
I = insoluble
(a) sodium sulfide + iron(II) nitrate
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Na+
Na+
1
S
Na2SO4
S
Equation
Na2S(aq) + FeSO4(aq) → Na2SO4(aq) + FeS(s)
S2SO425
S
Fe2+
Fe2+
5
S
FeS
I
SO42S22
I
(b) sodium hydroxide + barium nitrate
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Na+
Na+
1
S
NaNO3
OHNO31
S
Ba2+
NO3Ba2+
OH4
2
I
I
Ba(OH)2
Chapter 8 Solution Stoichiometry • MHR
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Product Solubility
S
S
Equation
NaOH(aq) + Ba(NO3)2(aq) → NR
(exception to rule)
(c) cesium phosphate + calcium bromide
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Cs+
Cs+
1
S
CsBr
S
Equation
2Cs3PO4(aq) + 3CaBr2(aq) → 6CsBr(aq) + Ca3(PO4)2(s)
PO43Br3
S
Ca2+
Ca2+
4
I
Ca3PO4
I
BrPO432
I
(d) sodium carbonate and sulfuric acid
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Na+
Na+
1
S
Na2SO4
S
Equation
Na2CO3(aq) + H2SO4(aq) → NR
CO32SO425
S
H+
H+
1
S
H2CO3
S
SO42CO322
I
(e) sodium nitrate and copper(II) sulfate
NO3SO425
S
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Na+
Na+
1
S
Na2SO4
S
Equation
NaNO3(aq) + CuSO4(aq) → NR
Cu2+
SO422+
Cu
NO35
1
S
S
Cu(NO3)2
S
(f) ammonium iodide and silver nitrate
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
NH4+ INH4+ NO31
1
S
S
NH4NO3
S
Equation
NH4I(aq) + AgNO3(aq) → NH4NO3(aq) + AgI(s)
Ag+
Ag+
2
I
AgI
S
NO3I3
S
(g) potassium carbonate and iron(II) nitrate
NO3CO322
I
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
K+
K+
1
S
KNO3
S
Equation
K2CO3(aq) + Fe(NO3)2(aq) → 2KNO3(aq) + FeCO3(s)
CO32NO31
S
Fe2+
Fe2+
5
S
FeCO3
I
(h) aluminum nitrate and sodium phosphate
Reactant Ions
Products Ions
Al3+
Al3+
NO3PO43-
Na+
Na+
PO43NO3-
Chapter 8 Solution Stoichiometry • MHR
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Guideline
Ion Solubility
Product
Product Solubility
5
S
AlPO4
I
2
I
1
1
S
S
NaNO3
I
Equation
Al(NO3)3(aq) + Na3PO4(aq) → 3NaNO3(aq) + AlPO4(s)
(i) potassium chloride and iron(II)
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
K+
K+
1
S
KNO3
S
ClNO31
S
Equation
KCl(aq) + Fe(NO3)2(aq) → NR
Fe2+
Fe2+
5
S
FeCl2
S
NO3Cl3
S
(j) ammonium sulfate and barium chloride
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
NH4+
NH4+
1
S
NH4Cl
S
Equation
(NH4)2SO4(aq) + BaCl2(aq) → 2NH4Cl(aq) + BaSO4(s)
SO42Cl3
S
Ba2+
Ba2+
4
I
BaSO4
I
ClSO425
S
(k) sodium sulfide and nickel(II) sulfate
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Na+
Na+
1
S
Na2SO4
S
Equation
Na2S(aq) + NiSO4(aq) → Na2SO4(aq) + NiS(s)
S2SO425
S
Ni2+
Ni2+
5
S
NiS
I
SO42S22
I
(l) lead(II) nitrate and potassium bromide
NO3Br3
S
K+
K+
1
S
KNO3
S
BrNO31
S
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Pb2+
Pb2+
2
I
PbBr2
I
Equation
Pb(NO3)2(aq) + 2KBr(aq) → 2KNO3(aq) + PbBr2(s)
Check Your Solution
Examine the final equation to see if the ionic compounds that are noted as being in
aqueous solution are soluble and the compounds noted as solid are insoluble.
Chapter 8 Solution Stoichiometry • MHR
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Solutions for Practice Problems
Student Textbook page 294
5. Problem
Mixing each pair of aqueous solutions results in a chemical reaction. Identify the
spectator ions. Then write the balanced net ionic equation.
(a) sodium carbonate and hydrochloric acid
(b) sulfuric acid and sodium hydroxide
What Is Required?
Identify the spectator ions and write a balanced net ionic equation.
What Is Given?
You know the chemical names of the compounds.
Plan Your Strategy
Write the chemical formula of each compound, then complete the chemical equation
for the reaction. Using the solubility guideline, note which product is soluble and
which is the precipitate. Replace the formulae of the soluble ionic compounds with
dissociated ions. Identify the ions that appear on both sides of the equation as
spectator ions. Rewrite the equation without the spectator ions.
Act on Your Strategy
(a) Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2CO3(aq)
H2CO3(aq) → CO2(g) + H2O()
2Na+(aq) + CO32-(aq) + 2H+(aq) + 2Cl-(aq) → 2Na+(aq) + 2Cl-(aq) + CO2(g) + H2O()
spectator ions are: Na+(aq) and Cl-(aq)
net ionic equation: 2H+(aq) + CO32-(aq) → CO2(g) + H2O()
(b) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O()
2H+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq) → 2Na+(aq) + SO42-(aq) + 2H2O()
spectator ions are: Na+(aq) and SO42-(aq)
net ionic equation: 2H+(aq) + OH-(aq) → H2O()
Check Your Solution
Examine the net ionic equation to confirm that no ions are common to both sides
and that the equation is balanced.
6. Problem
Identify the spectator ions for the reaction that takes place when each pair of aqueous
solutions is mixed. Then write the balanced net ionic equation.
(a) ammonium phosphate and zinc sulfate
(b) lithium carbonate and nitric acid
(c) sulfuric acid and barium hydroxide
What Is Required?
Identify the spectator ions and write a balanced net ionic equation.
What Is Given?
You know the chemical names of the compounds.
Chapter 8 Solution Stoichiometry • MHR
150
Plan Your Strategy
Write the chemical formula of each compound, then complete the chemical equation
for the reaction. Using the solubility guideline, note which product is soluble and
which is the precipitate. Replace the formulae of the soluble ionic compounds with
dissociated ions. Identify the ions that appear on both sides of the equation as spectator ions. Rewrite the equation without the spectator ions.
Act on Your Strategy
(a) 2(NH4)3PO4(aq) + 3ZnSO4(aq) → Zn3(PO4)2(s) + 3(NH4)2SO4(aq)
6NH4+(aq) + 2PO43-(aq) + 3Zn2+(aq) + 3SO42-(aq) → 3Zn2+(aq) + 2PO43-(aq)
+6NH4+(aq) + 3SO42-(aq)
spectator ions are: NH4+(aq) and SO42-(aq)
net ionic equation: 3Zn2+(aq) + 2PO43-(aq) → Zn3(PO4)2(s)
(b) Li2CO3(aq) + 2HNO3(aq) → 2LiNO3(aq) + H2O() + CO2(g)
2Li+(aq) + CO32-(aq) + 2H+(aq) + 2NO3-(aq) → 2Li+(aq) + 2NO3-(aq) + H2O()
+CO2(g)
spectator ions are: Li+(aq) and NO3-(aq)
net ionic equation: 2H+(aq) + CO32-(aq) → CO2(g) + H2O()
(c) H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O()
2H+(aq) + SO42-(aq) + Ba2+(aq) + 2OH-(aq) → BaSO4(s) + 2H2O()
spectator ions are: none
net ionic equation:
2H+(aq) + SO42-(aq) + Ba2+(aq) + 2OH-(aq) → BaSO4(s) + 2H2O()
Check Your Solution
Examine the net ionic equation to confirm that no ions are common to both sides
and that the equation is balanced.
Solutions for Practice Problems
Student Textbook page 300
7. Problem
Write dissociation equations for the following ionic compounds in water, and state
the concentration of the dissociated ions in each aqueous solution. Assume that each
compound dissolves completely.
(a) 1 mol/L lithium fluoride
(b) 0.5 mol/L sodium bromide
(c) 1.5 mol/L sodium sulfide
(d) 1 mol/L calcium nitrate
What is Required?
You must indicate the ions that form when the ionic compound dissociates and the
concentration of each ion in mol/L.
What is Given?
You know the name of the ionic compound and its concentration.
Plan Your Strategy
Write the chemical formula for each ionic compound. Recall that the chemical formula indicates the kind of number of ions in the compound. Refer to the student
textbook and periodic table for the correct formula and charge on the ions. The equation must be balance atomically and by charge. This means that the total charge on
the positive ion(s) must balance the total charge on the negative ions. The mol ratio
in the balanced dissociation equation will determine the concentration of each ion.
Chapter 8 Solution Stoichiometry • MHR
151
Act on Your Strategy
(a)
LiF(s)
Æ Li + ( aq) + F - (aq)
1 mol / L
1 mol / L
1 mol / L
(b)
NaBr(s)
Æ
Na + ( aq)
+
Br - (aq)
0.5 mol / L
0.5 mol / L
0.5 mol / L
(c)
Na 2 S(s)
Æ 2Na + ( aq) +
S2- (aq)
1.5 mol / L
3.0 mol / L
1.5 mol / L
(d)
Ca(NO 3 )2(s)
Æ
1.0 mol / L
Ca 2+ ( aq)
1.0 mol / L
+ 2NO 3
-
(aq)
2.0 mol / L
Check Your Solution
In each case the total charge on the positive ions numerically balances the total on the
negative ions. The concentration of the ions follows the mol ratio in the balanced dissociation equation.
8. Problem
Write dissociation equations for the following compounds, and state the concentration of the dissociated ions in each aqueous solution. Assume that each compound
dissolves completely.
(a) 1.75 mol/L ammonium chloride
(b) 0.25 mol/L ammonium sulfide
(c) 6 mol/L ammonium sulfate
(d) 2 mol/L silver nitrate
What is Required?
You must indicate the ions that form when the ionic compound dissociates and the
concentration of each ion in mol/L.
What is Given?
You know the name of the ionic compound and its concentration.
Plan Your Strategy
Write the chemical formula for each ionic compound. Recall that the chemical formula indicates the kind of number of ions in the compound. Refer to the student
textbook and periodic table for the correct formula and charge on the ions. The equation must be balance atomically and by charge. This means that the total charge on
the positive ion(s) must balance the total charge on the negative ions. The mol ratio
in the balanced dissociation equation will determine the concentration of each ion.
Act on Your Strategy
NH4 Cl(s)
Æ NH4 +( aq)
+
Cl -(aq)
1.75 mol/L
1.75 mol/L
1.75 mol/L
(a)
(b)
(NH4 )2 S(s) Æ 2NH4 +( aq) +
S2 -(aq)
0.25 mol/L
0.25 mol/L
0.50 mol/L
(c) (NH4 )2 SO4(s)
6.0 mol/L
(d)
Æ
2NH4 +( aq) + SO4 2 -(aq)
6.0 mol/L
12.0 mol/L
AgNO3(s) Æ
Ag +( aq)
+ NO3 -(aq)
2.0 mol/L
2.0 mol/L
2.0 mol/L
Chapter 8 Solution Stoichiometry • MHR
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Check Your Solution
In each case the total charge on the positive ions numerically balances the total on the
negative ions. The concentration of the ions follows the mol ratio in the balanced dissociation equation.
9. Problem
18.2 g of CaCl2 is dissolved in 300.0 mL of water to make a clear, colourless
solution. Give the dissociation equation for this physical change, and state the
concentration of the dissociated ions.
What is Required?
You must indicate the ions that form when the ionic compound dissociates and the
concentration of each ion in mol/L.
What is Given?
You know the name of the ionic compound and its concentration.
Plan Your Strategy
Write the chemical formula for calcium chloride and determine its molar mass. Use
m
n
the formula n =
to find the moles of calcium chloride. Use the formula C =
M
V
to determine the molar concentration. Write the dissociation equation and use the
mol ratio to indicate the concentration of each ion.
Act on Your Strategy
M for CaCl2 = 111 g/mol
1 mol
18.2 g CaCl2 ¥
= 0.164 mol CaCl2
111 g
n 0.164 mol
C= =
= 0.547 mol / L
V
0.300 L
CaCl 2(s)
Æ
Ca 2 +( aq)
+
2Cl -(aq)
0.547 mol/L
1.09 mol/L
0.547 mol/L
Check Your Solution
The equation is balanced atomically and by charge and the concentration of the ions
follows the mol ratio in this balanced dissociation equation.
10. Problem
Cobalt(II) chloride is a pale blue compound that dissolves easily in water. It is often
used to detect excess humidity in air, because the compound dissolves and turns pink.
(a) 14.2 g of cobalt(II) chloride is dissolved in 1.50 L of water. Determine the
concentration of CoCl2(aq).
(b) Write the dissociation equation for this change.
(c) Determine the concentration of all ions in the final solution.
What is Required?
You must indicate the ions that form when the ionic compound dissociates and the
concentration of each ion in mol/L.
What is Given?
You know the name of the ionic compound and its concentration.
Plan Your Strategy
Write the chemical formula for cobalt(II) chloride and determine its molar mass. Use
m
n
the formula n =
to find the moles of calcium chloride. Use the formula C =
M
V
to determine the molar concentration. Write the dissociation equation and use the
mol ratio to indicate the concentration of each ion.
Chapter 8 Solution Stoichiometry • MHR
153
Act on Your Strategy
(a) M for CoCl2 = 129.9 g/mol
14.2 g CoCl2 ¥ 1 mol = .109 mol CoCl2
129.8 g
n 0.109 mol
C =
=
= 0.0727 mol / L
V
1.5 0 L
(b) and (c)
CoCl 2(s)
Æ
Co2 +( aq)
+
2Cl -(aq)
0.0727 mol/L
0.145 mol/L
0.0727 mol/L
Check Your Solution
The equation is balanced atomically and by charge and the concentration of the ions
follows the mol ratio in this balanced dissociation equation.
Solutions for Practice Problems
Student Textbook page 302
11. Problem
Equal volumes of 0.150 mol/L lithium nitrate and 0.170 mol/L sodium nitrate are
mixed together. Determine the concentration of nitrate ions in the mixture.
What is Required?
You must calculate the concentration of the NO3–(aq) in a mixture of two ionic compounds.
What is Given?
The name and concentration of the two ionic compounds are given.
Plan Your Strategy
Write the chemical formula for each compound, write the dissociation equation and
use the mole ratio in this balanced equation to determine the concentration of the
NO3–(aq) . Let the volume of each solution = V. The number of moles of NO3–(aq)
can be calculated from n = C ¥ V. The final concentration of the NO3–(aq) can be
calculated as
Act on Your Strategy
LiNO3(s)
Æ
Li + (aq)
+
NO3 (aq)
0.150 mol / L
0.150 mol / L
0.150 mol / L
-
NaNO3(s)
Æ
Na + (aq)
+
NO3 (aq)
0.170 mol / L
0.170 mol / L
0.170 mol / L
total moles of ion
final concentration=
total volume
0.150 mol / L ¥ V + 0.170 mol / L ¥ V
final [NO3 _ (aq)] =
= 0.160 mol / L
2V
Check Your Solution
When mixing equal volumes of two aqueous solutions it is reasonable that the final
concentration will be an average of that of the two original concentrations. This
answer supports that reasoning.
Chapter 8 Solution Stoichiometry • MHR
154
12. Problem
Equal volumes of 1.25 mol/L magnesium sulfate and 0.130 mol/L ammonium sulfate
are mixed together. What is the concentration of sulfate ions in the mixture?
What is Required?
You must calculate the concentration of the SO42–(aq) in a mixture of two ionic compounds.
What is Given?
The name and concentration of the two ionic compounds are given.
Plan Your Strategy
Write the chemical formula for each compound, write the dissociation equation and
use the mole ratio in this balanced equation to determine the concentration of the
SO42–(aq). Let the volume of each solution = V. The number of moles of SO42–(aq)
can be calculated from n = C ¥ V. The final concentration of the SO42–(aq) can be
calculated as total moles of ion .
total volume
Act on Your Strategy
2MgSO4 (s) Æ
Mg 2 + (aq)
+ SO4 (aq)
1.25 mol / L
1.25 mol / L
1.25 mol / L
+
2-
(NH4 )2 SO4 (s) Æ
2NH4 (aq)
+
SO4 (aq)
0.130 mol / L
0.260 mol / L
0.130 mol / L
total moles of ion
final concentration =
total volume
1.25 mol / L ¥ V + 0.130 mol / L ¥ V
2final [SO4 (aq) ] =
= 0.690 mol / L
2V
Check Your Solution
When mixing equal volumes of two aqueous solutions it is reasonable that the final
concentration will be an average of that of the two original concentrations. This
answer supports that reasoning.
13. Problem
A solution is prepared by adding 25.0 mL of 0.025 mol/L sodium chloride with 35.0
mL of 0.050 mol/L barium chloride. What is the concentration of chloride ions in
the resulting solution, assuming that the volumes are additive?
What is Required?
You must calculate the concentration of the Cl–(aq) in a mixture of two ionic compounds.
What is Given?
The name and concentration of the two ionic compounds and the volume of each
solution are given.
Plan Your Strategy
Write the chemical formula for each compound, write the dissociation equation and
use the mole ratio in this balanced equation to determine the concentration of the
Cl–(aq). The number of moles of Cl–(aq) can be calculated from n = C ¥ V. The final
concentration of the Cl–(aq) can be calculated as total moles of ion
total volume
Chapter 8 Solution Stoichiometry • MHR
155
Act on Your Strategy
NaCl(s)
Æ
Na + (aq)
+
Cl - (aq)
0.025 mol / L
0.120 mol / L
0.025 mol / L
BaCl2 (s)
Æ
Ba 2 + (aq)
+
2Cl - (aq)
0.050 mol / L
0.10 mol / L
0.050 mol / L
total moles of ion
final concentration =
total volume
0.025 mol / L ¥ 0.0250 L + 0.10 mol / L ¥ 0.0350 L
final [Cl ] =
= 0.069 mol / L
0.060 L
Check Your Solution
The final concentration of the Cl–(aq) is expected to be a weighted average of the two
original concentrations. A larger volume and a higher concentration of Cl–(aq) comes
from the BaCl2 solution and therefore the final concentration of Cl–(aq) should be
closer to the original concentration of this solution. This is the case for the answer
obtained.
14. Problem
What is the concentration of nitrate ions in a solution that results by mixing 350 mL
of 0.275 mol/L potassium nitrate with 475 mL of 0.345 mol/L magnesium nitrate?
What is Required?
You must calculate the concentration of the NO3–(aq) in a mixture of two ionic compounds.
What is Given?
The name and concentration of the two ionic compounds and the volume of each
solution are given.
Plan Your Strategy
Write the chemical formula for each compound, write the dissociation equation and
use the mole ratio in this balanced equation to determine the concentration of the
NO3–(aq). The number of moles of NO3–(aq) can be calculated from n = C ¥ V. The
final concentration of the NO3–(aq) can be calculated as total moles of ion .
total volume
Act on Your Strategy
KNO3(s)
Æ
K + (aq)
+
NO3 (aq)
0.275 mol / L
0.275 mol / L
0.275 mol / L
Mg(NO3 )2 (s) Æ
Mg 2 + (aq)
+
2NO3 (aq)
0.345 mol / L
0.345 mol / L
0.690 mol / L
total moles of ion
final concentration =
total volume
0.275 mol / L ¥ 0.350 L + 0.690 mol / L ¥ 0.475 L
final [NO3 (aq) ] =
= 0.514 mol
0.825 L
Check Your Solution
The final concentration of the NO3–(aq) is expected to be a weighted average of the
two original concentrations. A larger volume and a higher concentration of NO3–(aq)
comes from the Mg(NO3)2 solution and therefore the final concentration of
NO3–(aq) should be closer to the original concentration of this solution. This is the
case for the answer obtained.
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Solutions for Practice Problems
Student Textbook page 304
15. Problem
Food manufacturers sometimes add calcium acetate to puddings and sweet sauces as a
thickening agent. What volume of 0.500 mol/L calcium acetate, Ca(CH3COO)2(aq),
contains 0.300 mol of acetate ions?
What Is Required?
Find the volume of 0.500 mol/L calcium acetate, Ca(CH3COO)2(aq), solution that
contains 0.300 mol of acetate ions?
What Is Given?
You know the concentration of the calcium acetate solution, the formula for calcium
acetate, and the required number of moles of acetate ion.
Plan Your Strategy
Write the equation for the dissociation of Ca(CH3COO)2. Determine the ratio of
moles CH3COO-(aq) : Ca(CH3COO)2 to calculate the moles of Ca(CH3COO)2(aq).
Rearrange the formula n = C ¥ V and calculate the volume of solution required.
Act on Your Strategy
Ca(CH3COO)2 → Ca2+(aq) + 2CH3COO-(aq)
0.300 CH3COO- ¥
V = Cn =
0.150 mol
0.500 mol/L
1 mol Ca(CH3COO)2
2 mol CH3COO -
= 0.150 mol Ca(CH3COO)2
= 0.300 L or 300 mL of Ca(CH 3COO)2(aq)
Check Your Solution
The final answer has the correct unit and number of significant figures. The answer
seems to be reasonable.
16. Problem
In a neutralization reaction, hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water:
HCl(aq) + NaOH(aq) Æ NaCl(aq) + H2O
What is the minimum volume of 0.676 mol/L HCl that is needed to combine exactly
with 22.7 mL of 0.385 mol/L aqueous sodium hydroxide?
What is Required?
You must find the volume of HCl that reacts with the NaOH.
What is Given?
You know a volume and concentration of NaOH and a concentration of HCl.
Plan Your Strategy
Calculate the mol of NaOH using n = C ¥ V. Use the mol ratio in the balanced
equation to calculate the mol of HCl that combines with this number of mol of
NaOH.
Use the formula V = n to calculate the volume of HCl.
C
Act on Your Strategy
mol NaOH = 0.385 mol/L ¥ 0.0227 L = 0.00874 mol
NaOH and HCl react 1:1
mol HCl required = 0.00874 mol = 0.676 mol/L ¥ V
V = 0.0129 L
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Check Your Answer
Since these reagents react in a mol ratio of 1:1, compare the amount of each using
rounded off values: amount HCl = 0.7 ¥ 0.013 = 0.009 mol;
mol NaOH = 0.4 ¥ 0.023 = 0.009.
The values are about the same and therefore the answer is reasonable.
17. Problem
Sodium sulfate and water result when sulfuric acid, H2SO4, reacts with sodium
hydroxide. What volume of 0.320 mol/L sulfuric acid is needed to react with
47.3 mL of 0.224 mol/L aqueous sodium hydroxide?
What is Required?
You must find the volume of H2SO4 that reacts with the NaOH.
What is Given?
You know a volume and concentration of NaOH and a concentration of HCl.
Plan Your Strategy
Calculate the mol of NaOH using n = C ¥ V. Write the balanced equation for this
reaction. Use the mol ratio in the balanced equation to calculate the mol of H2SO4
that combines with this number of mol of NaOH.
n
Use the formula V =
to calculate the volume of H2SO4.
C
Act on Your Strategy
H2SO4(aq) + 2 NaOH(aq) Æ Na2SO4(aq) + 2 H2O(l)
mol NaOH = 0.224 mol/L ¥ 0.0473 L = 0.0106 mol
0.0106 mol NaOH ¥ 1 mol H2 SO4 = 0.00530 mol H2SO4
2 mol NaOH
0.00530 mol H2SO4 = 0.320 mol/L ¥ V
V = 0.0166 L
Check Your Answer
Since these reagents react in a mol ratio of 1:2, compare the number of mol of each
using rounded off values. mol NaOH = 0.5 ¥ 0.22 = 0.11;
mol H2SO4 = 0.3 ¥ 0.02 = 0.06 The values are about in a ratio of 1:2 and therefore
our answer is reasonable.
18. Problem
Your stomach secretes hydrochloric acid to help you digest the food you have eaten.
If too much HCl is secreted, however, you may need to take an antacid to neutralize
the excess. One antacid product contains the compound magnesium hydroxide,
Mg(OH)2.
(a) Predict the reaction that takes place when magnesium hydroxide reacts with
hydrochloric acid (Hint: this is a double displacement reaction)
(b) Imagine that you are a chemical analyst testing the effectiveness of antacids. If
0.10 mol/L HCl serves as your model for stomach acid, how many litres will react
with an antacid that contains 0.10 g of magnesium hydroxide.
What is Required?
(a) Write the equation for the double displacement reaction between hydrochloric
acid and magnesium hydroxide.
(b) Find the volume of 0.10 mol/L HCl that will react with 0.10 g Mg(OH)2
What is Given?
You are given a mass of Mg(OH)2, a concentration of HCl and the periodic table to
find the molar mass of Mg(OH)2
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Plan Your Strategy
Write the balanced equation for this double displacement reaction. Calculate the
m
molar mass of Mg(OH)2 and convert 0.10 g to moles using n =
. Use the
M
mol ratio in the balanced equation to determine the mol of HCl that will react.
n
Determine the volume of HCl using V =
.
C
Act on Your Strategy
(a) Mg(OH)2(aq) + 2 HCl(aq) Æ MgCl2(aq) + 2H2O(l)
(b) M for Mg(OH)2 = 58.33 g/mol
0.10 g
mol Mg(OH)2 =
= 0.0017 mol
58.33 g/mol
0.0017 mol Mg(OH)2 ¥ 2 mol HCl = 0.0034 mol HCl
1 mol Mg()H)2
0.0034 mol
V=
= 0.034 L
0.10 mol/L
Therefore the volume of HCl that reacts is 34 mL.
Check Your Solution
Working backwards, using rounded off numbers, and a mol ratio of 1:2,
1
0.03 L ¥ 0.10 mol/L ¥
= 0.0015 mol. This is about the mol of Mg(OH)2 that
2 is reasonable.
have been given. The answer
Solutions for Practice Problems
Student Textbook pages 306–307
19. Problem
8.76 g of sodium sulfide is added to 350 mL of 0.250 mol/L lead(II) nitrate.
Calculate the maximum mass of precipitate that can form.
What Is Required?
You must calculate the mass of precipitate that forms when known amounts of solutions Na2S and Pb(NO3)2 react.
What Is Given?
You know a mass of Na2S and a volume and concentration of Pb(NO3)2 solution.
Plan Your Strategy
Find the molar mass of Na2S and calculate the moles of this compound. Use the formula n = C ¥ V to calculate the moles of Pb(NO3)2. Write the balanced equation for
the reaction between these two compounds and refer to the solubility guideline to
identify the insoluble product. Identify which of these two reactants is the limiting
reagent. Use the limiting reagent to calculate the moles of precipitate. Determine the
molar mass of the precipitate and convert the moles to grams.
Act on Your Strategy
molar mass of Na2S = 78.05 g/mol
8.76 g
mass
n = molar
= 78.05 g/mol = 0.112 mol Na2S
mass
n = C ¥ V = 0.250 mol/L ¥ 0.350 L = 0.0875 mol Pb(NO3)2
Na2S(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbS(s)
The precipitate is PbS.
Since the reactants react in a molar ratio of 1:1, the compound with the lesser number of moles will be the limiting reagent, i.e. the 0.0875 mol Pb(NO3)2. Also, since
the mol ratio of Pb(NO3)2 : PbS is 1:1, we can conclude that 0.0875 mol of
Pb(NO3)2 will produce 0.0875 mol of PbS.
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molar mass of PbS = 239.2 g/mol
239.2 g
mass of PbS - 0.0875 mol ¥ 1 mol = 20.9 g of PbS
Check Your Solution
The final answer has the correct unit and number of significant figures. This answer
seems reasonable based upon the mass of the sample.
20. Problem
25.0 mL of 0.400 mol/L Pb(NO3)2(aq) is mixed with 300 mL of 0.220 mol/L KI(aq).
What is the maximum mass of precipitate that can form?
What Is Required?
Find the mass of precipitate that will form when known volumes of two solutions of
given concentration are mixed?
What Is Given?
You are given the volumes, concentrations, names and formulae of two solutions that
are to be mixed.
Plan Your Strategy
Write the balanced equation for this reaction and identify the insoluble product. Use
the formula n = C ¥ V to calculate the number of moles of each reactant. Identify
the limiting reagent. Use the limiting reagent to calculate the moles of precipitate.
Find the molar mass of the precipitate and convert the moles to grams.
Act on Your Strategy
Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(aq) + PbI2(s)
moles Pb(NO3)2(aq) = n = C ¥ V = 0.400 mol/L ¥ 0.0250 L = 0.0100 mol
moles KI(aq) = n = C ¥ V = 0.220 mol/L ¥ 0.300 L = 0.0660 mol
From the balanced equation, Pb(NO3)2 : KI = 1 : 2.
2 mol KI
= 0.0200 mol KI
0.0100 mol Pb(NO3) ¥ 1 mol
Pb(NO3)2
But there are 0.0660 mol of KI available, therefore KI is in excess and Pb(NO3)2 is
the limiting reagent.
From the balanced equation, the molar ratio Pb(NO3)2 : PbI2 is 1:1.
Therefore, 0.0100 mol of PbI2 are produced.
molar mass of PbI2 = 461.0 g/mol
461.0 g
mass of PbI2 = 0.0100 mol ¥ mol = 4.61 g PbI2
Check Your Solution
The final answer has the correct unit and number of significant figures. This answer
seems reasonable based upon the mass of the sample.
21. Problem
Zoe mixes 15.0 mL of 0.250 mol/L aqueous sodium hydroxide with 20.0 mL of
0.400 mol/L aqueous aluminum nitrate.
(a) Write the chemical equation for the reaction.
(b) Calculate the maximum mass of precipitate that forms.
What is required?
You must write a balanced equation for this reaction and calculate the mass of precipitate that forms.
What is Given?
You have a volume and concentration for each of the reactants and the periodic table
to determine molar masses.
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Plan Your Strategy
This reaction is a double displacement reaction. Refer to the solubility guidelines in
the textbook to determine which product is the precipitate. Write the balanced equation. Calculate the mol of NaOH and Al(NO3)2 using n = C ¥ V and use the mole
ratio in the balanced equation to determine the limiting reagent. Use mol of limiting
reagent and the mol ratio in the balanced equation to calculate the mol of precipitate,
Al(OH)3. Use the formula m = n ¥ M to calculate the mass of Al(OH)3.
Act on Your Strategy
(a) 3 NaOH(aq) + Al(NO3)2(aq) Æ 3 NaNO3(aq) + Al(OH)3(s)
(b) mol NaOH(aq) = 0.250 mol/L ¥ 0.0150 L = 0.00375 mol
mol Al(NO3)2(aq) = 0.400 mol/L ¥ 0.0200 L = 0.00800 mol
NaOH(aq) and Al(NO3)2(aq) react in a ratio of 3:1.
0.00375 mol NaOH ¥ 1 mol Al(NO 3 )2 = 0.00125 mol Al(NO3)2(aq)
3 mol NaOH
Since 0.00800 mol of Al(NO3)2(aq) are given, and 0.00125 mol are required to
react with all of the NaOH, Al(NO3)2(aq) is in excess and NaOH is the limiting
reagent.
1 mol Al(OH)3
0.00375 mol NaOH ¥
= 0.00125 mol Al(OH)3(s)
3 mol NaOH
molar mass Al(OH)3 = 77.98 g/mol
mass of Al(OH)3 = 0.00125 mol ¥ 77.98 g/mol = 0.0975 g
Therefore the mass of precipitate, Al(OH)3, is 0.0975 g
Check Your Solution
Working backwards from the mass of precipitate, verify that the mol of limiting
reagent corresponds to the data given in the question. the answer has the correct unit
and number of significant figures and seems reasonable.
22. Problem
In an experiment, Kendra mixed 40.0 mL of 0.552 mol/L lead(II) nitrate with 50.0
mL of 1.22 mol/L hydrochloric acid. A white precipitate formed which Kendra filtered and dried. She determined the mass to be 5.012 g.
(a) What is the formula of the precipitate that formed?
(b) Which is the limiting reactant?
(c) What is the theoretical yield?
(d) Calculate the percentage yield.
What is Required?
You must determine what compound precipitates and write its formula. Write a balanced equation for the reaction and determine which is the limiting reactant. Finally,
you must calculate the theoretical and percentage yields.
What is Given?
You know the volume and concentration of each reactant and can use the periodic
table to determine molar masses.
Plan Your Strategy
(a) This reaction is a double displacement reaction. Refer to the solubility guidelines
in the student textbook to determine which product is the precipitate.
(b) Write the balanced equation.
(c) Calculate the mol of Pb(NO3)2(aq) and HCl(aq) using n
= C ¥ V and use the mole
ratio in the balanced equation to determine the limiting reagent. Use mol of
limiting reagent and the mol ratio in the balanced equation to calculate the mol of
Chapter 8 Solution Stoichiometry • MHR
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precipitate, PbCl2(s) Use the formula m = n ¥ M to calculate the mass of PbCl2(s).
This is the theoretical yield.
(d) Calculate the percentage yield using the formula
actual yield
percentage yield =
¥ 100%
theoretical yield
Act on Your Strategy
(a) The precipitate is PbCl2
(b) 2 HCl(aq) + Pb(NO3)2(aq) Æ 2 HNO3(aq) + PbCl2(s)
(c) mol HCl(aq) = 1.22 mol/L ¥ 0.0500 L = 0.0610 mol
mol Pb(NO3)2(aq) = 0.552 mol/L ¥ 0.0400 L = 0.0221 mol
0.0610 mol HCl(aq) ¥ 1 mol PbCl2 = 0.0305 mol PbCl2(s)
1 mol Pb(NO3 )2
1 mol PbCl 2 = 0.0221 mol PbCl
2(s)
1 mol Pb(NO 3 )2
Since the Pb(NO3)2(aq) gives a smaller amount of PbCl2(s), Pb(NO3)2(aq) is the
limiting reagent.
molar mass of PbCl2(s) = 278.1 g/mol
mass of PbCl2(s) = 0.0221 mol ¥ 278.1 g/mol = 6.15 g PbCl2(s)
Therefore the theoretical yield of PbCl2(s) is 6.15 g.
0.0221 mol Pb(NO3)2(aq) ¥
actual yield
5.012 g
¥ 100% =
100% = 81.5 %
theoretical yield
6.15 g
Therefore the percentage yield is 81.5%
(d) percentage yield =
Check Your Solution
Work backwards to verify that this yield matches the data in the question. The
answer seems reasonable and has the correct number of significant figures and units.
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