CHEMISTRY 11
Chapter 9
Aqueous Solutions
Solutions for Practice Problems
Section 9.1 Making Predictions About Solubility
Student Textbook page 335
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 on page 334
of the text 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
lead(II)
chloride
zinc oxide
Formula
PbCl2
ZnO
Ions
Guideline
Ion Solubility
2+
2
insoluble
−
Cl
3
soluble
Zn2+
5
soluble
2−
2
insoluble
2
insoluble
1
soluble
Pb
O
silver acetate
AgCH3COO
Ag+
CH3COO
−
Salt Solubility
INSOLUBLE
INSOLUBLE
SOLUBLE
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.
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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 on page 334
of the text 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
potassium
nitrate
KNO3
Lithium
carbonate
Li2CO3
lead(II) oxide
Ions
Guideline
Ion Solubility
1
soluble
NO3
1
soluble
Li+
1
soluble
CO3
2
insoluble
Pb2+
2
insoluble
2
insoluble
+
K
−
2−
PbO
2−
O
Salt Solubility
SOLUBLE
SOLUBLE
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 on page 334
of the text 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
2+
Ca
Guideline
Ion Solubility
4
insoluble
CO3
2−
2
insoluble
Mg2+
5
soluble
2−
5
soluble
5
soluble
2
insoluble
SO4
Al3+
PO4
3−
Salt Solubility
INSOLUBLE
SOLUBLE
INSOLUBLE
Check Your Solution
Check the predictions against another reference. These results are correct.
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Section 9.2 Reactions in Aqueous Solutions
Student Textbook page 339
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 on page 334 of the 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)
S2−
SO42−
5
S
Fe2+
Fe2+
5
S
FeS
I
SO42−
S2−
2
I
(b) sodium hydroxide + barium nitrate
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Na+
Na+
1
S
NaNO3
S
OH−
NO3−
1
S
Ba2+
NO3−
Ba2+
OH−
4
2
I
I
Ba(OH)2
S
(exception to rule)
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Equation
NaOH(aq) + Ba(NO3)2(aq) → NR
(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)
PO43−
Br−
3
S
Ca2+
Ca2+
4
I
Ca3PO4
I
Br−
PO43−
2
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
CO32−
SO42−
5
S
H+
H+
1
S
H2CO3
S
SO42−
CO32−
2
I
(e) sodium nitrate and copper(II) sulfate
NO3−
SO42−
5
S
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
Na+
Na+
1
S
Na2SO4
S
Equation
NaNO3(aq) + CuSO4(aq) → NR
Cu2+
SO42−
2+
Cu
NO3−
5
1
S
S
Cu(NO3)2
S
(f) ammonium iodide and silver nitrate
Reactant Ions
Products Ions
Guideline
Ion Solubility
Product
Product Solubility
NH4+ I−
NH4+ NO3−
1
1
S
S
NH4NO3
S
Equation
NH4I(aq) + AgNO3(aq) → NH4NO3(aq) + AgI(s)
Ag+
Ag+
2
I
AgI
S
NO3−
I−
3
S
(g) potassium carbonate and iron(II) nitrate
NO3−
CO32−
2
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)
CO32−
NO3−
1
S
Fe2+
Fe2+
5
S
FeCO3
I
(h) aluminum nitrate and sodium phosphate
Reactant Ions
Products Ions
Guideline
Al3+
Al3+
5
NO3−
PO43−
2
Na+
Na+
1
PO43−
NO3−
1
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CHEMISTRY 11
Ion Solubility
Product
Product Solubility
S
AlPO4
I
I
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
Cl−
NO3−
1
S
Equation
KCl(aq) + Fe(NO3)2(aq) → NR
Fe2+
Fe2+
5
S
FeCl2
S
NO3−
Cl−
3
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)
SO42−
Cl−
3
S
Ba2+
Ba2+
4
I
BaSO4
I
Cl−
SO42−
5
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)
S2−
SO42−
5
S
Ni2+
Ni2+
5
S
NiS
I
SO42−
S2−
2
I
(l) lead(II) nitrate and potassium bromide
NO3−
Br−
3
S
K+
K+
1
S
KNO3
S
Br−
NO3−
1
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.
Solutions for Practice Problems
Student Textbook page 343
5. Problem
Mixing each pair of aqueous solutions results in a chemical reaction. Identify the
spectator ions. Then write the balanced net ionic equation.
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CHEMISTRY 11
(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.
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)
+
2−
spectator ions are: NH4 (aq) and SO4 (aq)
net ionic equation: 3Zn2+(aq) + 2PO43−(aq) → Zn3(PO4)2(s)
(b) Li2CO3(aq) + 2HNO3(aq) → 2LiNO3(aq) + H2O() + CO2(g)
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CHEMISTRY 11
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.
Section 9.3 Stoichiometry in Solution Chemistry
Student Textbook page 352
7. 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=
n
C
=
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(CH3COO)2(aq)
Check Your Solution
The final answer has the correct unit and number of significant figures. The answer
seems to be reasonable.
8. Problem
Ammonium phosphate can be used as a fertilizer. 6.0 g of ammonium phosphate is
dissolved in sufficient water to produce 300 mL of solution. What are the concentrations (in mol/L) of the ammonium ions and phosphate ions present?
What Is Required?
Find the concentrations of the NH4+(aq) and PO43−(aq) ions in solution.
What Is Given?
You know the mass of ammonium phosphate and the final volume of the solution.
Chapter 9 Aqueous Solutions • MHR
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CHEMISTRY 11
Plan Your Strategy
Calculate the molar mass of (NH4)3PO4 and convert 6.0 g of this compound to
moles. Calculate the concentration of (NH4)3PO4 in mol/L.Write the balanced equation for the dissociation of (NH4)3PO4 and use the mole ratios in this equation to
determine the concentration of each of the ions.
Act on Your Strategy
molar mass of (NH4)3PO4 = 149.1 g/mol
1 mol
= 0.040 mol (NH4)3PO4
6.0 g (NH4)3PO4 × 149.1
g
mol
= 0.13 mol/L
concentration of (NH4)3PO4 = 0.040
0.300 L
+
3−
(NH4)3PO4(s) → 3NH4 (aq) + PO4 (aq)
→
= 0.39 mol NH4+(aq)
moles NH4+(aq) = 0.13 mol/L (NH4)3PO4
→
= 0.13 mol PO43−(aq)
moles PO43−(aq) = 0.13 mol (NH4)3PO4
Check Your Solution
The final answer has the correct unit and number of significant figures. This answer
seems reasonable.
9. Problem
An aqueous solution of a certain salt contains chloride ions. A sample of this solution
was made by dissolving 17.59 g of the salt in a 1 L volumetric flask. Ten 25.00 mL of
the solution was treated with excess silver nitrate. The precipitate, AgCl, was filtered
and dried. If the mass of this dry precipitate was 47.35 g, what was the mass percent
of chloride ions in the solution?
What Is Required?
Determine the mass percent of chloride ion in a salt solution.
What Is Given?
You know the original mass of salt used to prepare the solution. Also, you know the
mass of AgCl that precipitates from a 25.00 mL sample of the solution reacts with an
excess of AgNO3 .
Plan Your Strategy
Determine the molar mass of AgCl and convert 0.4735 g of this compound to
moles. From the mol ratio of Cl− in AgCl, determine the moles of chloride ion in the
precipitate. All of the chloride from the 25.00 mL sample of salt solution is in this
AgCl. Calculate the moles of Cl− in 1.000 L of solution and use the molar mass of
Cl− to convert this to grams. Use this mass and the original mass of the salt solution
to determine the mass percent of Cl− in the solution.
Act on Your Strategy
molar mass AgCl = 143.4 g/mol
mass of Cl− in precipitate = mass of Cl− in 25.00 mL sample
1 mol
= 0.4735 g AgCl 143.4
= 0.003302 g Cl−
g
35.45 g Cl−
1000 mL
× 1 mol =
25.00 mL
mass of Cl−
× 100%
mass of sample
4.682 g
× 100% = 26.62%
17.59 g
mass of Cl− per 1.000 L = 0.003302 mol Cl−
mass percent of Cl− in original solution =
=
4.682 g
Check Your Solution
The final answer has the correct unit and number of significant figures. This answer
seems reasonable.
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10. Problem
The active ingredient in some rat poisons is thallium(I) sulfate, Tl2SO4 . A chemist
takes a 500 mg sample of thallium(I) sulfate and adds potassium iodide, to precipitate
yellow thallium(I) iodide. When the precipitate is dried, its mass is 200 mg. What is
the mass percent of Tl2SO4 in the rat poison?
What Is Required?
Calculate the mass percent of Tl2SO4 in 500 mg sample of rat poison.
What Is Given?
You know the original mass of the rat poison, the mass of a precipitate of TlI and the
chemical formula of thallium sulfate.
Plan Your Strategy
All of the thallium in the precipitate TlI was in the original sample of rat poison.
Determine the molar mass of TlI and calculate the moles of Tl in the 200 mg of the
precipitate. Write the balance equation for the reaction between Tl2SO4 and KI and
use the mole ratio in this equation to calculate the moles of Tl2SO4 . Find the molar
mass of Tl2SO4 and calculate the mass of this compound. Use this mass and the mass
of the sample of rat poison to determine the mass percent of Tl2SO4 in the rat
poison.
Act on Your Strategy
molar mass of TlI = 331.28 g/mol
Tl2SO4(aq) + 2KI(aq) → 2TlI(s) + K2SO4(aq)
1 mol TlI
×
moles Tl2SO4 = 0.200 g TlI × 331.28
g TlI
3.019 × 10−3
1 mol Tl2SO4
2 mol TlI
= 3.019 × 10−3 mol Tl2SO4
504.83 g
mol Tl2SO4 × 1 mol = 0.152 g Tl2SO4
mass percent Tl2SO4 in sample =
=
mass of Tl2SO4
× 100%
mass of sample
0.152 g
× 100% = 30.5%
0.500 g
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.
Solutions for Practice Problems
Student Textbook page 355
11. 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
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CHEMISTRY 11
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.
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.
12. 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.
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13. Problem
A student 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?
For the reaction between two solutions, write the balanced equation and determine
the mass of precipitate that will form.
What Is Given?
You are given the volumes, concentrations, names, and chemical formulae of the two
solutions.
Plan Your Strategy
(a) This reaction is a double displacement reaction that follows the form:
AB + CD → CB + AD . Refer to Table 9.1 General Solubility Guidelines and
determine which product is a precipitate. Write the chemical formula for each
compound and balance the equation.
(b) Rewrite the balanced equation showing each soluble ionic compound dissociated
into its ions. Write the net ionic equation for the reaction. Calculate the number
of moles of each reactant ion. Identify the limiting reagent. Use the limiting
reagent to determine the number of moles of precipitate that form. Calculate the
molar mass of the precipitate and convert the number of moles to grams.
(a) 3NaOH(aq) + Al(NO3)3(aq) → 3NaNO3(aq) + Al(OH)3(s)
(b) 3Na+(aq) + 3OH−(aq) + Al3+(aq) + 3NO3−(aq) → 3Na+(aq) + 3NO3−(aq) + Al(OH)3(s)
net ionic equation: Al3+(aq) + 3OH−(aq) → Al(OH)3(s)
Al(OH)3 is the precipitate.
1 mol Al3+
× 0.400 mol/L Al(NO3)3 × 0.0200 L = 0.00800 mol
moles Al3+ = 1 mol
Al(NO )
3 3
−
mol OH
× 0.250 mol/L × 0.0150 L = 0.00375 mol
moles OH− = 11mol
NaOH
3+ :
−
Since Al OH is 1:3, OH− will produce less Al(OH)3 and is the limiting
reagent.
Al(OH)3
× 0.00375 mol OH− = 0.00125 mol
moles of Al(OH)3 = 1 3mol
mol OH−
mass of Al(OH)3 = 0.00125mol ×
77.98 g
1 mol
= 0.0975 g
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.
Chapter 9 Aqueous Solutions • MHR
164