CHAPTER 4
Chemical Reactions
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
4.1. a. According to Table 4.1, all compounds that contain sodium, Na+, are soluble. Thus, NaBr is
soluble in water.
b. According to Table 4.1, most compounds that contain hydroxides, OH, are insoluble in
water. However, Ba(OH)2 is listed as one of the exceptions to this rule, so it is soluble in
water.
c. Calcium carbonate is CaCO3. According to Table 4.1, most compounds that contain
carbonate, CO32−, are insoluble. CaCO3 is not one of the exceptions, so it is insoluble in
water.
4.2. a. The problem states that HNO3 is a strong electrolyte, but Mg(OH)2 is a solid, so retain its
formula. On the product side, Mg(NO3)2 is a soluble ionic compound, but water is a
nonelectrolyte, so retain its formula. The resulting complete ionic equation is
2H+(aq) + 2NO3−(aq) + Mg(OH)2(s) 2H2O(l) + Mg2+(aq) + 2NO3−(aq)
The corresponding net ionic equation is
2H+(aq) + Mg(OH)2(s) 2H2O(l) + Mg2+(aq)
b. Both reactants are soluble ionic compounds, and on the product side, NaNO3 is also a
soluble ionic compound. PbSO4 is a solid, so retain its formula. The resulting complete
ionic equation is
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + SO42−(aq) PbSO4(s) + 2Na+(aq) + 2NO3−(aq)
The corresponding net ionic equation is
Pb2+(aq) + SO42−(aq) PbSO4(s)
4.3. The formulas of the compounds are NaI and Pb(C2H3O2)2. Exchanging anions, you get sodium
acetate, NaC2H3O2, and lead(II) iodide, PbI2. The equation for the exchange reaction is
NaI + Pb(C2H3O2)2 NaC2H3O2 + PbI2
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, 114 Chapter 4: Chemical Reactions
From Table 4.1, you see that NaI is soluble, Pb(C2H3O2)2 is soluble, NaC2H3O2 is soluble, and
PbI2 is insoluble. Thus, lead(II) iodide precipitates. The balanced molecular equation with phase
labels is
Pb(C2H3O2)2(aq) + 2NaI(aq) PbI2(s) + 2NaC2H3O2(aq)
To get the net ionic equation, you write the soluble ionic compounds as ions and cancel the
spectator ions, (C2H3O2− and Na+). The final result is
Pb2+(aq) + 2I−(aq) PbI2(s)
4.4. a. H3PO4 is not listed as a strong acid in Table 4.3, so it is a weak acid.
b. Hypochlorous acid, HClO, is not one of the strong acids listed in Table 4.3, so we assume
that HClO is a weak acid.
c. As noted in Table 4.3, HClO4 is a strong acid.
d. As noted in Table 4.3, Sr(OH)2 is a strong base.
4.5. The salt consists of the cation from the base (Li+) and the anion from the acid (CN−); its formula
is LiCN. You will need to add H2O as a product to complete and balance the molecular equation:
HCN(aq) + LiOH(aq) LiCN(aq) + H2O(l)
Note that LiOH (a strong base) and LiCN (a soluble ionic substance) are strong electrolytes; HCN
is a weak electrolyte (it is not one of the strong acids in Table 4.3). After eliminating the spectator
ions (Li+ and CN−), the net ionic equation is
HCN(aq) + OH−(aq) H2O(l) + CN−(aq)
4.6. The first step in the neutralization is described by the following molecular equation:
H2SO4(aq) + KOH(aq) KHSO4(aq) + H2O(l)
The corresponding net ionic equation is
H+(aq) + OH−(aq) H2O(l)
The reaction of the acid salt KHSO4 is given by the following molecular equation:
KHSO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)
The corresponding net ionic equation is
HSO4−(aq) + OH−(aq) H2O(l) + SO42−(aq)
4.7. First, write the molecular equation for the exchange reaction, noting that the products of the
reaction would be soluble Ca(NO3)2 and H2CO3. The carbonic acid decomposes to water and
carbon dioxide gas. The molecular equation for the process is
CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O(l) + CO2(g)
The corresponding net ionic equation is
CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) + CO2(g)
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, Chapter 4: Chemical Reactions 115
4.8. a. For potassium dichromate, K2Cr2O7,
2 (oxidation number of K) + 2 (oxidation number of Cr) + 7 (oxidation number
of O) = 0
For oxygen, the oxidation number is −2 (rule 3), and for potassium ion, the oxidation
number is +1 (rule 2)
[2 (+1)] + 2 (oxidation number of Cr) + [7 (−2)] = 0
Therefore,
2 oxidation number of Cr = − [2 (+1)] − [7 (−2)] = +12
or, oxidation number of Cr = +6.
b. For the permanganate ion, MnO4−,
(Oxidation number of Mn) + 4 (oxidation number of O) = −1
For oxygen, the oxidation number is −2 (rule 3).
(oxidation number of Mn) + [4 (−2)] = −1
Therefore,
Oxidation number of Mn = −1 − [4 (−2)] = +7
4.9. Identify the oxidation states of the elements.
0 0 +2 -1
Ca + Cl2 CaCl2
Break the reaction into two half-reactions, making sure that both mass and charge are balanced.
Ca Ca2+ + 2e−
Cl2 + 2e− 2Cl−
Since each half-reaction has two electrons, it is not necessary to multiply the reactions by any
factors to cancel them out. Adding the two half-reactions together and canceling out the electrons,
you get
Ca(s) + Cl2(g) CaCl2(s)
4.10. Convert mass of NaCl (molar mass, 58.44 g) to moles of NaCl. Then divide moles of solute by
liters of solution. Note that 25.0 mL = 0.0250 L.
1 mol NaCl
0.0678 g NaCl = 1.160 10−3 mol NaCl
58.44 g NaCl
1.160 103 mol NaCl
Molarity = = 0.04641 = 0.0464 M
0.0250 L soln
4.11. Convert grams of NaCl (molar mass, 58.44 g) to moles NaCl and then to volume of NaCl
solution.
1 mol NaCl 1 L solution 1000 mL
0.0958 g NaCl = 10.06 = 10.1 mL NaCl
58.44 g NaCl 0.163 mol NaCl 1L
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, 116 Chapter 4: Chemical Reactions
4.12. One (1) liter of solution is equivalent to 0.15 mol NaCl. The amount of NaCl in 50.0 mL of
solution is
1L 0.15 mol NaCl
50.0 mL = 0.00750 mol NaCl
1000 mL 1 L soln
Convert to grams using the molar mass of NaCl (58.44 g/mol).
58.4 g NaCl
0.00750 mol NaCl = 0.438 = 0.44 g NaCl
1 mol NaCl
4.13. Use the rearranged version of the dilution formula from the text to calculate the initial volume of
1.5 M sulfuric acid required:
M f Vf 0.18 M 100.0 mL
Vi = = = 12.0 = 12 mL
Mi 1.5 M
4.14. There are two different reactions taking place in forming the CaC2O4 (molar mass 128.10 g/mol)
precipitate. These are
CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)
CaCl2(aq) + Na2C2O4(aq) CaC2O4(s) + 2NaCl(aq)
The overall stoichiometry of the reactions is one mol CaCO3/one mol CaC2O4. Also note that
each CaCO3 contains one Ca atom, so this gives an overall conversion factor of one mol Ca/one
mol CaC2O4.
The mass of Ca can now be calculated.
1 mol CaC2 O4 1 mol Ca 40.08 g Ca
0.1402 g CaC2O4 = 0.043866 g Ca
128.10 g CaC2O 4 1 mol CaC2 O4 1 mol Ca
Now, calculate the percentage of calcium in the 128.3 mg (0.1283 g) limestone:
0.043866 g Ca
100% = 34.190 = 34.19%
0.1283 g limestone
4.15. Convert the volume of Na3PO4 to moles using the molarity of Na3PO4. Note that 45.7 mL =
0.0457 L.
0.265 mol Na 3 PO4
0.0457 L Na3PO4 = 0.01211 mol Na3PO4
1L
Finally, calculate the amount of NiSO4 required to react with this amount of Na3PO4:
3 mol NiSO4 1 L NiSO4
0.1211 mol Na3PO4 = 0.04844 L (48.4 mL)
2 mol Na 3 PO4 0.375 mol NiSO4
4.16. Convert the volume of NaOH solution (0.0391 L) to moles NaOH (from the molarity of NaOH).
Then convert moles NaOH to moles HC2H3O2 (from the chemical equation). Finally, convert
moles of HC2H3O2 (molar mass 60.05 g/mol) to grams HC2H3O2.
0.108 mol NaOH 1 mol HC2 H3O2 60.05 g HC2 H3O2
0.0391 L NaOH = 0.25359 g
1L 1 mol NaOH 1 mol HC2 H3O2
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chemical Reactions
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
4.1. a. According to Table 4.1, all compounds that contain sodium, Na+, are soluble. Thus, NaBr is
soluble in water.
b. According to Table 4.1, most compounds that contain hydroxides, OH, are insoluble in
water. However, Ba(OH)2 is listed as one of the exceptions to this rule, so it is soluble in
water.
c. Calcium carbonate is CaCO3. According to Table 4.1, most compounds that contain
carbonate, CO32−, are insoluble. CaCO3 is not one of the exceptions, so it is insoluble in
water.
4.2. a. The problem states that HNO3 is a strong electrolyte, but Mg(OH)2 is a solid, so retain its
formula. On the product side, Mg(NO3)2 is a soluble ionic compound, but water is a
nonelectrolyte, so retain its formula. The resulting complete ionic equation is
2H+(aq) + 2NO3−(aq) + Mg(OH)2(s) 2H2O(l) + Mg2+(aq) + 2NO3−(aq)
The corresponding net ionic equation is
2H+(aq) + Mg(OH)2(s) 2H2O(l) + Mg2+(aq)
b. Both reactants are soluble ionic compounds, and on the product side, NaNO3 is also a
soluble ionic compound. PbSO4 is a solid, so retain its formula. The resulting complete
ionic equation is
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + SO42−(aq) PbSO4(s) + 2Na+(aq) + 2NO3−(aq)
The corresponding net ionic equation is
Pb2+(aq) + SO42−(aq) PbSO4(s)
4.3. The formulas of the compounds are NaI and Pb(C2H3O2)2. Exchanging anions, you get sodium
acetate, NaC2H3O2, and lead(II) iodide, PbI2. The equation for the exchange reaction is
NaI + Pb(C2H3O2)2 NaC2H3O2 + PbI2
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, 114 Chapter 4: Chemical Reactions
From Table 4.1, you see that NaI is soluble, Pb(C2H3O2)2 is soluble, NaC2H3O2 is soluble, and
PbI2 is insoluble. Thus, lead(II) iodide precipitates. The balanced molecular equation with phase
labels is
Pb(C2H3O2)2(aq) + 2NaI(aq) PbI2(s) + 2NaC2H3O2(aq)
To get the net ionic equation, you write the soluble ionic compounds as ions and cancel the
spectator ions, (C2H3O2− and Na+). The final result is
Pb2+(aq) + 2I−(aq) PbI2(s)
4.4. a. H3PO4 is not listed as a strong acid in Table 4.3, so it is a weak acid.
b. Hypochlorous acid, HClO, is not one of the strong acids listed in Table 4.3, so we assume
that HClO is a weak acid.
c. As noted in Table 4.3, HClO4 is a strong acid.
d. As noted in Table 4.3, Sr(OH)2 is a strong base.
4.5. The salt consists of the cation from the base (Li+) and the anion from the acid (CN−); its formula
is LiCN. You will need to add H2O as a product to complete and balance the molecular equation:
HCN(aq) + LiOH(aq) LiCN(aq) + H2O(l)
Note that LiOH (a strong base) and LiCN (a soluble ionic substance) are strong electrolytes; HCN
is a weak electrolyte (it is not one of the strong acids in Table 4.3). After eliminating the spectator
ions (Li+ and CN−), the net ionic equation is
HCN(aq) + OH−(aq) H2O(l) + CN−(aq)
4.6. The first step in the neutralization is described by the following molecular equation:
H2SO4(aq) + KOH(aq) KHSO4(aq) + H2O(l)
The corresponding net ionic equation is
H+(aq) + OH−(aq) H2O(l)
The reaction of the acid salt KHSO4 is given by the following molecular equation:
KHSO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)
The corresponding net ionic equation is
HSO4−(aq) + OH−(aq) H2O(l) + SO42−(aq)
4.7. First, write the molecular equation for the exchange reaction, noting that the products of the
reaction would be soluble Ca(NO3)2 and H2CO3. The carbonic acid decomposes to water and
carbon dioxide gas. The molecular equation for the process is
CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O(l) + CO2(g)
The corresponding net ionic equation is
CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) + CO2(g)
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, Chapter 4: Chemical Reactions 115
4.8. a. For potassium dichromate, K2Cr2O7,
2 (oxidation number of K) + 2 (oxidation number of Cr) + 7 (oxidation number
of O) = 0
For oxygen, the oxidation number is −2 (rule 3), and for potassium ion, the oxidation
number is +1 (rule 2)
[2 (+1)] + 2 (oxidation number of Cr) + [7 (−2)] = 0
Therefore,
2 oxidation number of Cr = − [2 (+1)] − [7 (−2)] = +12
or, oxidation number of Cr = +6.
b. For the permanganate ion, MnO4−,
(Oxidation number of Mn) + 4 (oxidation number of O) = −1
For oxygen, the oxidation number is −2 (rule 3).
(oxidation number of Mn) + [4 (−2)] = −1
Therefore,
Oxidation number of Mn = −1 − [4 (−2)] = +7
4.9. Identify the oxidation states of the elements.
0 0 +2 -1
Ca + Cl2 CaCl2
Break the reaction into two half-reactions, making sure that both mass and charge are balanced.
Ca Ca2+ + 2e−
Cl2 + 2e− 2Cl−
Since each half-reaction has two electrons, it is not necessary to multiply the reactions by any
factors to cancel them out. Adding the two half-reactions together and canceling out the electrons,
you get
Ca(s) + Cl2(g) CaCl2(s)
4.10. Convert mass of NaCl (molar mass, 58.44 g) to moles of NaCl. Then divide moles of solute by
liters of solution. Note that 25.0 mL = 0.0250 L.
1 mol NaCl
0.0678 g NaCl = 1.160 10−3 mol NaCl
58.44 g NaCl
1.160 103 mol NaCl
Molarity = = 0.04641 = 0.0464 M
0.0250 L soln
4.11. Convert grams of NaCl (molar mass, 58.44 g) to moles NaCl and then to volume of NaCl
solution.
1 mol NaCl 1 L solution 1000 mL
0.0958 g NaCl = 10.06 = 10.1 mL NaCl
58.44 g NaCl 0.163 mol NaCl 1L
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, 116 Chapter 4: Chemical Reactions
4.12. One (1) liter of solution is equivalent to 0.15 mol NaCl. The amount of NaCl in 50.0 mL of
solution is
1L 0.15 mol NaCl
50.0 mL = 0.00750 mol NaCl
1000 mL 1 L soln
Convert to grams using the molar mass of NaCl (58.44 g/mol).
58.4 g NaCl
0.00750 mol NaCl = 0.438 = 0.44 g NaCl
1 mol NaCl
4.13. Use the rearranged version of the dilution formula from the text to calculate the initial volume of
1.5 M sulfuric acid required:
M f Vf 0.18 M 100.0 mL
Vi = = = 12.0 = 12 mL
Mi 1.5 M
4.14. There are two different reactions taking place in forming the CaC2O4 (molar mass 128.10 g/mol)
precipitate. These are
CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)
CaCl2(aq) + Na2C2O4(aq) CaC2O4(s) + 2NaCl(aq)
The overall stoichiometry of the reactions is one mol CaCO3/one mol CaC2O4. Also note that
each CaCO3 contains one Ca atom, so this gives an overall conversion factor of one mol Ca/one
mol CaC2O4.
The mass of Ca can now be calculated.
1 mol CaC2 O4 1 mol Ca 40.08 g Ca
0.1402 g CaC2O4 = 0.043866 g Ca
128.10 g CaC2O 4 1 mol CaC2 O4 1 mol Ca
Now, calculate the percentage of calcium in the 128.3 mg (0.1283 g) limestone:
0.043866 g Ca
100% = 34.190 = 34.19%
0.1283 g limestone
4.15. Convert the volume of Na3PO4 to moles using the molarity of Na3PO4. Note that 45.7 mL =
0.0457 L.
0.265 mol Na 3 PO4
0.0457 L Na3PO4 = 0.01211 mol Na3PO4
1L
Finally, calculate the amount of NiSO4 required to react with this amount of Na3PO4:
3 mol NiSO4 1 L NiSO4
0.1211 mol Na3PO4 = 0.04844 L (48.4 mL)
2 mol Na 3 PO4 0.375 mol NiSO4
4.16. Convert the volume of NaOH solution (0.0391 L) to moles NaOH (from the molarity of NaOH).
Then convert moles NaOH to moles HC2H3O2 (from the chemical equation). Finally, convert
moles of HC2H3O2 (molar mass 60.05 g/mol) to grams HC2H3O2.
0.108 mol NaOH 1 mol HC2 H3O2 60.05 g HC2 H3O2
0.0391 L NaOH = 0.25359 g
1L 1 mol NaOH 1 mol HC2 H3O2
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