Chapter 4: 11, 16, 18, 20, 22, 24, 30, 34, 36, 40, 42, 44, 46, 48, 50, 56, 60, 62, 64, 76, 82, 94
11.
a.
b.
c.
d.
e.
f.
g.
h.
i.
NaBr
MgCl2
Al(NO3)3
(NH4)2SO4
HI
FeSO4
KMnO4
HClO4
NH4C2H3O2
Na+ + Br–
Mg2+ + 2 Cl–
Al3+ + 3 NO3–
2 NH4+ + SO42–
H+ + I–
Fe2+ + SO42–
K+ + MnO4–
H+ + ClO4–
NH4+ + C2H3O2–
1 mole NaCl
58.44 g NaCl
= 0.2815 M NaCl
1.000 L
1 g/ KIO 3
1 mole KIO 3
853.5 mg/ KIO 3 x
x
3
214.0 g/ KIO 3
10 mg/ KIO 3
= 1.595 x 10 − 2 M KIO 3
1L
250.0 mL/ x
10 3 mL/
1 mole Fe
1 mole Fe 3+
0.4508 g/ Fe x
x
55.85 g/ Fe
1 mole Fe
= 1.614 x 10 − 2 M Fe 3+
1L
500.0 mL/ x
10 3 mL/
16.45 g NaCl x
16.
a.
b.
c.
18.
a.
Ca(NO3)2
Ca2+ + 2 NO3–
0.100 mole Ca(NO 3 ) 2
= 1.00 M Ca(NO 3 ) 2
0.100 L
1.00 mole Ca(NO 3 ) 2
1 mole Ca 2+
x
= 1.00 M Ca 2+
L
1 mole Ca(NO 3 ) 2
−
2 mole NO 3
1.00 mole Ca(NO 3 ) 2
−
= 2.00 M NO 3
x
1 mole Ca(NO 3 ) 2
L
b.
2 Na+ + SO42–
Na2SO4
2.5 mole Na 2 SO 4
= 2.0 M Na 2 SO 4
1.25 L
2.0 mole Na 2 SO 4
2 mole Na +
x
= 4.0 M Na +
L
1 mole Na 2 SO 4
2−
1 mole SO 4
2.0 mole Na 2 SO 4
2−
= 2.0 M SO 4
x
1 mole Na 2 SO 4
L
c.
NH4+ + Cl–
1 mole NH 4 Cl
5.00 g NH 4 Cl x
53.49 g NH 4 Cl
= 0.187 M NH 4 Cl
0.5000 L
+
0.187 mole NH 4 Cl
1 mole NH 4
+
x
= 0.187 M NH 4
L
1 mole NH 4 Cl
NH4Cl
0.187 mole NH 4 Cl
1 mole Cl −
x
= 0.187 M Cl −
L
1 mole NH 4 Cl
d.
3 K+ + PO43–
1 mole K 3 PO 4
1.00 g K 3 PO 4 x
212.27 g K 3 PO 4
= 0.0188 M K 3 PO 4
0.2500 L
0.0188 mole K 3 PO 4
3 mole K +
x
= 0.0565 M K +
L
1 mole K 3 PO 4
K3PO4
3−
0.0188 mole K 3 PO 4
1 mole PO 4
3−
x
= 0.0188 M PO 4
L
1 mole K 3 PO 4
20.
0.100 mole NaOH
x 0.100 L/ = 0.0100 mole NaOH
L/
NaOH
Na+ + OH–
1 mole Na +
= 0.0100 mole Na +
1 mole NaOH
1 mole OH −
0.0100 mole NaOH x
= 0.0100 mole OH −
1 mole NaOH
0.0100 mole NaOH x
0.0100 mole + 0.0100 mole = 0.0200 mole ions
0.200 mole BaCl 2
x 0.0500 L/ = 0.0100 mole BaCl 2
L/
BaCl2
Ba2+ + 2 Cl–
0.0100 mole BaCl 2
1 mole Ba 2+
x
= 0.0100 mole Ba 2+
1 mole BaCl 2
0.0100 mole BaCl 2
2 mole Cl −
x
= 0.0200 mole Cl −
1 mole BaCl 2
0.0100 mole + 0.0200 mole = 0.0300 mole ions
0.150 mole Na 3 PO 4
x 0.0750 L/ = 0.01125 mole Na 3 PO 4 2
L/
Na3PO4
3 Na+ + PO43–
0.01125 mole Na 3 PO 4 x
3 mole Na +
= 0.03375 mole Na +
1 mole Na 3 PO 4
0.01125 mole Na 3 PO 4 x
1 mole PO 4
3−
= 0.01125 mole PO 4
1 mole Na 3 PO 4
3−
0.03375 mole + 0.01125 mole = 0.0450 mole ions
22.
0.400 mole NaOH
40.00 g NaOH
x 0.2500 L/ x
= 4.00 g NaOH
L/
1 mole NaOH
24.
a.
b.
c.
0.50 mole H 2 SO 4
1L
= 0.028 L
x 1.00 L/ x
18 mole H 2 SO 4
L/
28 mL diluted to 1.00 L
0.50 mole HCl
1L
x 1.00 L/ x
= 0.042 L
L/
12 mole HCl
42 mL diluted to 1.00 L
237.71 g NiCl 2 ⋅ 6 H 2 O
1 mole NiCl 2 ⋅ 6 H 2 O
0.50 mole NiCl 2
x
x 1.00 L/ x
1 mole NiCl 2 ⋅ 6 H 2 O
1 mole NiCl 2
L/
= 1.2 x 10 2 g NiCl 2 ⋅ 6 H 2 O
1.2 x 102 g NiCl2 . 6 H2O dissolved in enough water to give to 1.00 L
d.
e.
0.50 mole HNO 3
1L
= 0.031 L
x 1.00 L/ x
16 mole HNO 3
L/
31 mL diluted to 1.00 L
0.50 mole Na 2 CO 3
105.99 g Na 2 CO 3
x 1.00 L/ x
= 53 g Na 2 CO 3
L/
1 mole Na 2 CO 3
53 g Na2CO3 dissolved in enough water to give to 1.00 L
30.
34.
a.
FeSO4(aq) + KCl(aq)
b.
2 Al(NO3)3(aq) + 3 Ba(OH)2(aq)
c.
CaCl2(aq) + Na2SO4(aq)
CaSO4(s) + 2 NaCl(aq)
d.
K2S(aq) + Ni(NO3)2(aq)
NiS(s) + 2 KNO3(aq)
a.
2 AgNO3(aq) + BaCl2(aq)
Ag+(aq) + Cl–(aq)
b.
FeSO4(aq) + K2S(aq)
Fe2+(aq) + S2– (aq)
c.
NaOH(aq) + K2SO4(aq)
d.
Hg2(NO3)2(aq) + CaCl2(aq)
Hg22+(aq) + 2 Cl–(aq)
36.
a.
CoCl3(aq) + 3 NaOH(aq)
Co3+(aq) + 3 OH–(aq)
b.
2 Al(OH)3(s) + 3 Ba(NO3)2(aq)
2 AgCl(s) + Ba(NO3)2(aq)
AgCl(s)
FeS(s) + K2SO4(aq)
FeS(s)
NR
Hg2Cl2(s) + Ca(NO3)2(aq)
Hg2Cl2(s)
Co(OH)3(s) + 3 NaCl(aq)
Co(OH)3(s)
2 AgNO3(aq) + (NH4)2CO3(aq)
2 Ag+(aq) + CO32–(aq)
40.
NR
c.
CuSO4(aq) + HgCl2(aq)
d.
Sr(NO3)2(aq) + KI(aq)
3 Pb(NO3)2(aq) + 2 Na3PO4(aq)
Ag2CO3(s) + 2 NH4NO3(aq)
Ag2CO3(s)
NR
NR
Pb3(PO4)2(s) + 6 NaNO3(aq)
0.250 mole Pb(NO 3 ) 2
2 mole Na 3 PO 4
1L
x 0.150 L/ x
x
0.100 mole Na 3 PO 4
L/
3 mole Pb(NO 3 ) 2
= 0.250 L = 250. mL Na3PO4 solution
42.
3 BaCl2(aq) + Fe2(SO4)3(aq)
3 BaSO4(s) + 2 FeCl3(aq)
0.100 mole BaCl 2
3 mole BaSO 4
233.4 g BaSO 4
x 0.100 L/ x
x
L/
3 mole BaCl 2
1 mole BaSO 4
= 2.33 g BaSO4
0.100 mole Fe 2 (SO 4 ) 3
3 mole BaSO 4
233.4 g BaSO 4
x 0.100 L/ x
x
L/
1 mole Fe 2 (SO 4 ) 3
1 mole BaSO 4
= 7.00 g BaSO4
44.
2 AgNO3(aq) + CaCl2(aq)
2 AgCl(s) + Ca(NO3)2(aq)
0.20 mole AgNO 3
2 mole AgCl
143.4 g AgCl
x 0.1000 L/ x
x
L/
2 mole AgNO 3
1 mole AgCl
= 2.9 g AgCl
0.15 mole CaCl 2
2 mole AgCl
143.4 g AgCl
x 0.1000 L/ x
x
L/
1 mole CaCl 2
1 mole AgCl
= 4.3 g AgCl
Ag+:
AgNO3 limiting; 0.00 M Ag+
Ca2+:
0.20 mole AgNO 3
1 mole CaCl 2
x 0.1000 L/ x
= 0.010 mole CaCl 2 used up
L/
2 mole AgNO 3
0.15 mole CaCl 2
x 0.1000 L/ − 0.010 mole CaCl 2 = 0.005 mole CaCl 2 remaining
L/
0.20 mole AgNO 3
1 mole Ca(NO 3 ) 2
x 0.1000 L/ x
= 0.010 mole Ca(NO 3 ) 2 formed
L/
2 mole AgNO 3
There is 1 mole of Ca2+/ 1 mole of each compound, therefore the total Ca2+ is:
0.005 mole + 0.010 mole = 0.015 mole (Note: this is identical to moles of CaCl2
originally added. Ca2+ does not precipitate in this reaction.)
0.015 mole Ca 2+
= 0.075 M Ca2+
0.1000 L + 0.1000 L
Cl -:
0.15 mole CaCl 2
x 0.1000 L/ − 0.010 mole CaCl 2 = 0.005 mole CaCl 2 remaining
L/
2 mole Cl −
0.005 mole CaCl 2 x
= 0.010 mole Cl - not in AgCl
1 mole CaCl 2
0.010 mole Cl −
= 0.050 M Cl–
0.1000 L + 0.1000 L
NO3 -:
−
0.20 mole AgNO 3
1 mole Ca(NO 3 ) 2
2 mole NO 3
−
x 0.1000 L/ x
x
= 0.020 mole NO 3
L/
2 mole AgNO 3
1 mole Ca(NO 3 ) 2
−
0.020 mole NO 3
= 0.10 M NO3–
0.1000 L + 0.1000 L
46.
a.
3 HNO3(aq) + Al(OH)3(s)
Al(NO3)3(aq) + 3 H2O(l)
3 H+ + 3 NO3– + Al(OH)3
3 H+ + Al(OH)3
b.
Al3+ + 3 NO3– + 3 H2O
Al3+ + 3 H2O
HC2H3O2(aq) + KOH(aq)
KC2H3O2(aq) + H2O(l)
HC2H3O2 + K+ + OH–
K+ + C2H3O2– + H2O
HC2H3O2 + OH–
c.
C2H3O2– + H2O
Ca(OH)2(aq) + 2 HCl(aq)
CaCl2(aq) + 2 H2O(l)
Ca2+ + 2 OH– + 2 H+ + 2 Cl–
OH– + H+
48.
a.
Ca2+ + 2 Cl– + 2 H2O
H2O
AgOH(s) + HBr(aq)
AgBr(s) + H2O(l)
AgOH + H+ + Br–
AgBr + H2O
Same
b.
Sr(OH)2(aq) + 2 HI(aq)
SrI2(aq) + 2 H2O(l)
Sr2+ + 2 OH– + 2 H+ + 2 I–
OH - + H+
c.
H2O
Fe(OH)3(s) + 3 HNO3(aq)
Fe(OH)3 + 3 H+ + 3 NO3–
Fe(OH)3 + 3 H+
Sr2+ + 2 I– + 2 H2O
Fe(NO3)3(aq) + 3 H2O(l)
Fe3+ + 3 NO3– + 3 H2O
Fe3+ + 3 H2O
50.
0.200 mole HCL
x 0.02500 L/ = 0.00500 mole HCl
L/
a.
HCl(aq) + NaOH(aq)
0.00500 mole HCl x
b.
BaCl2(aq) + 2 H2O(l)
1 mole Ba(OH) 2
1L
= 0.0500 L = 50.0 mL
x
0.0500 mole Ba(OH) 2
2 mole HCl
HCl(aq) + KOH(aq)
0.00500 mole HCl x
56.
1 mole NaOH
1L
x
= 0.0500 L = 50.0 mL
1 mole HCl
0.100 mole NaOH
Ba(OH)2(aq) + 2 HCl(aq)
0.00500 mole HCl x
c.
NaCl(aq) + H2O(l)
KCl(aq) + H2O(l)
1 mole KOH
1L
x
= 0.0200 L = 20.0 mL
1 mole HCl
0.250 mole KOH
NaOH(aq) + KHC8H4O4(aq)
or
NaOH(aq) + KHP(aq)
NaKC8H4O4 (aq) + H2O(l)
NaKP(aq) + H2O(l)
1 mole KHP
1 mole NaOH
x
204.22 g/ KHP
1 mole KHP
= 0.02590 M NaOH
1L
20.46 mL/ x
10 3 mL/
0.1082 g/ KHP x
58.
60.
a.
UO22+
+6 -2
b.
As2O3
+3 -2
c.
NaBiO3
+1 +5 -2
d.
As4
0
e.
HAsO2
+1 +3 –2
f.
Mg2P2O7
+2 +5 –2
g.
Na2S2O3
+1 +2 –2
h.
Hg2Cl2
+1 –1
i.
Ca(NO3)2
+2 +5 –2
a.
Li3N
+1 –3
b.
NH3
–3 +1
c.
N2H4
–2 +1
d.
NO
+2 –2
e.
N2O
+1 –2
f.
NO2
+4 –2
g.
NO2–
+3 –2
h.
NO3–
+5 –2
i.
N2
0
62.
a.
Cu + 2 Ag+
0
+1
2 Ag+
Cu
Cu
Ag+
Cu
Ag+
b.
2 Ag + Cu2+
0
+2
2 Ag
Cu2+
reduction
oxidation
reducing agent
oxidizing agent
substance oxidized
substance reduced
HCl + NH3
+1 –1 –3 +1
NH4Cl
–3 +1 –1
Not redox
c.
SiCl4 + 2 H2O
+4 –1 +1 –2
4 HCl + SiO2
+1 -1 +4 –2
Not redox
d.
SiCl4 + 2 Mg
+4 –1
0
Si
MgCl2
SiCl4
Mg
Mg
SiCl4
Mg
SiCl4
e.
2 MgCl2 + Si
+2 –1
0
reduction
oxidation
reducing agent
oxidizing agent
substance oxidized
substance reduced
Al(OH)4–
+3 –2 +1
AlO2– + 2 H2O
+3 –2
+1 –2
Not redox
64.
a.
Cu
NO3Cu
NO3-
Cu2+
NO
Cu2+
NO + 2 H2O
Cu
Cu2+
4 H+ + NO3–
Cu
NO + 2 H2O
Cu2+ + 2 e-
3 e- + 4 H+ + NO3-
NO + 2 H2O
Cu2+ + 2 e-)
3 (Cu
2 (3 e- + 4 H+ + NO3–
NO + 2 H2O)
6 e- + 3 Cu + 8 H+ + 2 NO3b.
Cr2O72Cl–
Cr3+
Cl2
Cr2O72–
2 Cl–
Cr2O72–
2 Cl–
2 Cr3+
Cl2
2 Cr3+ + 7 H2O
Cl2
14 H+ + Cr2O72–
2 Cl -
2 Cr3+ + 7 H2O
Cl2
6 e- + 14 H+ + Cr2O72–
2 Cl–
2 Cr3+ + 7 H2O
Cl2 + 2 e-
6 e- + 14 H+ + Cr2O72–
3(2 Cl–
2 NO + 4 H2O + 3 Cu2+ + 6 e-
Cl2 + 2 e-)
2 Cr3+ + 7 H2O
6 e- + 6 Cl– + 14 H+ + Cr2O72–
c.
Pb + H2SO4
2 Cr3+ + 7 H2O + 3 Cl2 + 6 e-
PbSO4
PbO2 + H2SO4
PbSO4
Pb + H2SO4
PbSO4
PbO2 + H2SO4
PbSO4 + 2 H2O
PbSO4 + 2 H+
Pb + H2SO4
2 H+ + PbO2 + H2SO4
PbSO4 + 2 H2O
PbSO4 + 2 H+ + 2 e-
Pb + H2SO4
2 e- + 2 H+ + PbO2 + H2SO4
2 e- + Pb + 2 H+ + 2 H2SO4 + PbO2
d.
Mn2+
NaBiO3
Mn2+
NaBiO3
4 H2O + Mn2+
NaBiO3
4 H2O + Mn2+
6 H+ + NaBiO3
PbSO4 + 2 H2O
2 PbSO4 + 2 H2O + 2 H+ + 2 e-
MnO4–
Bi3+
MnO4–
Bi3+ + Na+
MnO4–
Bi3+ + Na+ + 3 H2O
MnO4– + 8 H+
Bi3+ + Na+ + 3 H2O
4 H2O + Mn2+
MnO4– + 8 H+ + 5 e-
2e- + 6 H+ + NaBiO3
Bi3+ + Na+ + 3 H2O
MnO4– + 8 H+ + 5 e-)
2(4 H2O + Mn2+
5(2e- + 6 H+ + NaBiO3
Bi3+ + Na+ + 3 H2O)
14
10 e- + 8 H2O + 2 Mn2+ + 30 H+ + 5 NaBiO3
7
+ Bi3+ + 15 H2O + 5 Na+ + 10 e2 Mn2+ + 14 H+ + 5 NaBiO3
e.
Zn
AsH3
Zn2+
H3AsO4
Zn
AsH3 + 4 H2O
Zn2+
8 H+ + H3AsO4
Zn
AsH3 + 4 H2O
Zn2+ + 2 e-
8 e- + 8 H+ + H3AsO4
4 (Zn
2 MnO4– + Bi3+ + 7 H2O + 5 Na+
Zn2+
H3AsO4
Zn
2 MnO4– + 16 H+
AsH3 + 4 H2O
Zn2+ + 2 e-)
8 e- + 8 H+ + H3AsO4
8 e- + 4 Zn + 8 H+ + H3AsO4
AsH3 + 4 H2O
AsH3 + 4 H2O + 4 Zn2+ + 8 e-
76.
0.5032 g/ BaSO 4 x
1 mole/ BaSO 4
1 mole/ Sac
183.19 g Sac
x
x
= 0.3950 g Sac
233.4 g/ BaSO 4 1 mole/ BaSO 4
1 mole/ Sac
0.3950 g Sac
= 0.03950 g Sac Tablet
10 Tablets
0.5894 g
= 0.05894 g Tablet
10 Tablets
0.03950 g Sac
x 100 = 67.01 % saccharin
0.05894 g Tablets
82.
a.
Al(s) + 3 HCl(aq)
2 Al(s) + 6 HCl(aq)
0
+1 –1
Al
H+
b.
c.
94.
CO2
H2O
Cu(s) + 2 Ag+(aq)
0
+1
Cu
Ag+
CS2(l) + 2 H2S(g)
+4 –2
+1 –2
CS2
oxidation
CS2 and H2S reduction
C3H8(g) + 5 O2(g)
–2 2/3 +1
0
C3H8
O2
d.
Al3+ oxidation
H2 reduction
CH4(g) + 3 S(s)
–4 +1
0
CH4
S
AlCl3(aq) + 3/2 H2(g)
2 AlCl3(aq) + 3 H2(g)
+3 –1
0
Cu2+
Ag
3 CO2(g) + 4 H2O(l)
+4 –2
+1 –2
oxidation
reduction
2 Ag(s) + Cu2+(aq)
0
+2
oxidation
reduction
Assume 100 g of compound:
53.66 g/ C x
4.09 g/ H x
1 mole C
= 4.468 mole C/2.641 = 1.692
12.01 g/ C
1 mole H
= 4.06 mole H/2.641 = 1.537
1.008 g/ H
The %O comes from 100 - (53.66 + 4.09) = 42.25
42.25 g/ O x
1 mole O
= 2.641 mole O/2.641 = 1
16.00 g/ O
The factor needed to give a whole number is not obvious, but the "formula weight" can be
determined:
"FW" = 1.692 x 12.011 + 1.537 x 1.008 + 1 x 16.00 = 37.87 g/mole
The titration gives the actual formula weight:
NaOH + HCarm
NaCarm + H2O
0.0406 mole NaOH
1 mole HCarm
x 0.01802 L/ x
= 7.32 x 10 −4 mole HCarm
L/
1 mole NaOH
0.3602 g HCarm
= 492 g mole
7.32 x 10 − 4 mole Hcarm
492 g/mole
= 12.00 ≈ 13
37.87 g/mole
53.66 g/ C x
4.09 g/ H x
42.25 g/ O x
1 mole C
= 4.468 mole C/2.641 = 1.692 x 13 = 21.996 ≈ 22
12.01 g/ C
1 mole H
= 4.06 mole H/2.641 = 1.537 x 13 = 19.981 ≈ 20
1.008 g/ H
1 mole O
= 2.641 mole O/2.641 = 1 x 13 = 13
16.00 g/ O
C22H20O13 or HC22H19O13