7.
Practice Problems
54.0 g H2O/151 g x 100 = 35.8 % H2O
1.
.0250 L Sr..x .800 mol Sr..x1 mol Cu..x1 L Cu..=.0267 L Cu..
1 L Sr..
1 mol Sr.. .750 mol Cu..
b.
3.21 g Hydrate x 64.2 g BeC2O4 = 2.06 g BeC2O4
100 g Hydrate
c. (1)
2.
.0353 L Mg(Cl)2 x .125 mol Mg.. = 4.41 x 10-3 mol Mg..
1 L Mg..
4.41 x 10-3 mol Mg.. x 1 mol Sr.. = 4.41 x 10-3 mol Sr..
1 mol Mg..
4.41 x 10-3 mol Sr../0.0548 L = 0.0805 M Sr(OH)2
0.0178 L x 0.0150 mol/1 L = 0.000267 mol KMnO4
(2)
0.000267 mol Mn... x 5 mol Be... x 96.9 g = 0.0647 g
2 mol Mn... 1 mol Be...
(3)
3.
0.0232 L SO42- x 0.150 mol SO42- x 1 mol Ba2+ x 137 g Ba = .477 g Ba
1 L SO421 mol SO42- 1 mol Ba
0.477 g Ba/9.00 g Ore x 100 = 5.30 %
4.
a.
a.
.
0.200 L FeCl2 x 0.600 mol/1 L = 0.120 mol FeCl3
0.150 L BaS x 0.500 mol/1 L = 0.0750 mol BaS
b.
0.120 mol FeCl2 x 1 mol Fe2S3 = 0.0600 mol Fe2S3
2 mol FeCl3
0.0750 mol BaS x 1 mol Fe2S3 = 0.0250 mol Fe2S3
3 mol BaS
c.
0.0250 mol Fe2S3 x 208 g Fe2S3 = 5.20 g Fe2S3
1 mol Fe2S3
d.
0.120 mol FeCl3  0.120 mol Fe3+, 0.360 mol Cl0.0750 mol BaS 0.0750 mol Ba2+, 0.0750 mol S2e.
0.0647 g BeC2O4/0.345 g sample x 100 = 18.7 %
8.
.0353 L Cr2O72- x .050 mol Cr2O72- x 6 mol Fe2+ = .0106 mol
1 L Cr2O721 mol Cr2O720.0106 mol Fe x 55.8 g Fe = .591 g Fe x 100 = 39.4 %
1 mol Fe 1.50 g Ore
9.
0.0250 mol Fe2S3 0.0500 mol Fe3+, 0.0750 mol S2f.
[Fe3+]: (0.120 – 0.0500)mol/0.350 L = 0.200 M
[Cl-]: 0.360 mol/0.350 L = 1.03 M
[Ba2+] = 0.0750 mol/.350L = 0.214 M
[S2-]: (0.0750 – 0.0750) mol/0.350 L = 0 M
5. a.
0.300 L x 0.500 mol/L = 0.150 mol Ca(NO3)2
0.200 L x 0.500 mol/L = 0.100 mol Li2CO3
b.
.150 mol Ca(NO3)2 x 1 mol CaCO3 = .150 mol CaCO3
1 mol Ca(NO3)2
.100 mol Li2CO3 x 1 mol CaCO3 = .100 mol CaCO3
1 mol Li2CO3
 Li2CO3
c.
0.100 mol CaCO3 x 100 g CaCO3 = 10.0 g CaCO3
1 mol CaCO3
d.
0.150 mol Ca(NO3)2 0.150 mol Ca2+, 0.300 mol NO30.100 mol Li2CO3 0.200 mol Li2+, 0.100 mol CO32e.
0.100 mol CaCO3 0.100 mol
Ca2+,
0.100 mol CO3
Anion
O2-
Ca2+
H-
CaH2
Calcium hydride
Al3+
Br-
AlBr3
Aluminum bromide
Ag+
S2-
Ag2S
Silver sulfide
Zn2+
I-
ZnI2
Zinc iodide
Pb3(PO4)4
Lead(IV) phosphate
CuSO4
Copper(II) sulfate
Cr(NO3)3
Chromium nitrate
Pb(IV)4+
PO4
Cu(II)2+
SO42-
Formula
FeO
3-
-
Name
Iron(II) oxide
Cr3+
NO3
Co(III)3+
Cl-
CoCl3
Cobalt(III) chloride
Sn(IV)4+
MnO4-
Sn(MnO4)4
Tin(IV) permanganate
Hg2(I)2+
CO32-
Hg2CO3
Mercury(I) carbonate
Na2Cr2O7
Sodium dichromate
Na+
Cr2O7
2-
Mg2+
C2H3O2
Mg(C2H3O2)2
Magnesium acetate
Ni2+
CrO42-
NiCrO4
Nickel chromate
2-
K2CO3
Potassium carbonate
K+
CO3
-
10.
2-
f.
[Ca2+]: (0.150 – 0.100)mol/0.500 L = 0.100 M
[NO3-]: 0.300 mol/0.500 L = 0.600 M
[Li+]: 0.200 mol/0.500 L = 0.400 M
[CO32-]: (0.100 – 0.100)mol/0.500 L = 0 M
6.
.0157 L Sr(OH)2 x 3.00 mol Sr(OH)2 = 0.0471 mol Sr(OH)2
1 L Sr(OH)2
0.0471 mol Sr(OH)2 x 2 mol HCl = 0.0942 mol HCl
1 mol Sr(OH)2
0.0942 mol HCl/0.0250 L = 3.77 M HCl
Cation
Fe(II)2+
Name
Ammonium carbonate
Cation
NH4+
Anion
CO32-
Formula
(NH4)2CO3
Aluminum acetate
Al3+
C2H3O2-
Al(C2H3O2)3
Silver sulfate
Ag+
SO4
Lead(IV) chlorite
Pb4+
ClO2-
Pb(ClO2)4
Copper(II) nitrite
Cu2+
NO2-
Cu(NO2)2
Sodium cyanide
Na+
CN-
NaCN
Copper(I) fluoride
Cu+
F-
CuF
Manganese(IV) oxide
Mn4+
O2-
MnO2
Iron(III) sulfite
Fe3+
SO32-
Magnesium dichromate
Mg2+
Cr2O7
Mercury(II) hydroxide
Hg2+
OH-
Potassium phosphate
K+
Sodium bicarbonate
Na+
PO4
2-
2-
Fe2(SO3)3
MgCr2O7
Hg(OH)2
3-
HCO3
Ag2SO4
-
K3PO4
NaHCO3
11.
IO4periodate
IO3iodate
IO2iodite
IOhypoiodite
2 Al(s) + 6 H+  2 Al3+ + 3 H2(g)
12.
1 MgCl2(aq) + 2 KOH(aq)  1 Mg(OH)2(s) + 2 KCl(aq)
2 NaBr(aq) + 1 Cl2(g)  2 NaCl(aq) + 1 Br2(l)
Mg2+ + 2 OH-  Mg(OH)2(s)
2 Br- + Cl2(g)  2 Cl- + Br2(l)
1 BaCl2(aq) + 1 Na2SO4(aq)  1 BaSO4(s) + 2 NaCl(aq)
Ba2+ + SO42-  BaSO4(s)
1 Ni(NO3)2(aq) + 1 Na2S(aq)  1 NiS(s) + 2 NaNO3(aq)
Ni2+ + S2-  NiS(s)
B
.025 L x 0.20 mol BrO3-/L x 3 mol Br2/1 mol BrO3- = 0.015
.030 L x 0.45 mol Br-/L x 3 mol Br2/5 mol Br- = 0.0081
2.
13.
sodium carbonate +
barium nitrate
mercury(I) sulfate +
ammonium chloride
magnesium nitrate +
sodium hydroxide
lead(II) nitrate +
potassium bromide
Practice Multiple Choice
1.
Ba2+ + CO32-  BaCO3(s)
Hg22+ + 2 Cl-  Hg2Cl2(s)
Mg2+ + 2 OH-  Mg(OH)2(s)
Pb2+ + 2 Br-  PbBr2(s)
A
0.010 L x 6 mol HNO3/L x 1 mol NO/4 mol HNO3 = 0.015
0.10 mol Ag x 1 mol/3 mol Ag = 0.033  0.015 mol NO
3.
B
1.0 L x 0.25 mol KOH x 1 mol SO2 x 64 g SO2 = 8.0 g
1L
2 mol KOH 1 mol SO2
4.
A
0.3 L x 0.2 mol Fe(NO3)3/1L = 0.06 mol x 3 = 0.18 mol
0.3 L x 1 mol NO3-/1L = 0.3 mol needed  0.06 mol Ba...
5.
14.
AgCl(s) + 2 NH3  [Ag(NH3)2] + Cl
+
-
B
Al(OH)3(s) + OH-  [Al(OH4)]-
6.
Cu(OH)2(s) + 4 Cl-  [CuCl4]2- + 2 OH+

15. a.
Co2+ + 4 Cl-  [Co(Cl)4]2b.
AgCl(s) + Cl-  [AgCl2]+
16. a.
Solid AgCl goes into solution as complex ion.
AgCl(s) + 2 NH3(aq)  Ag(NH3)2+ + Clb.
Light blue Cu2+ ion changes to dark blue complex.
(light) Cu2+ + 4 NH3(aq)  Cu(NH3)42+ (dark)
17.
H2SO4
H2SO3
HClO4
Fe3+
SCN-
[FeSCN]2+
C
A
D
D
C
nitric acid
nitrous acid
12.
HCl
HNO3
HNO2
A
 H2O
2 HBr(aq) + 1 Sr(OH)2(aq)  2 H2O(l) + 1 Sr(Br)2(aq)
2 HNO3(aq) + 1 SrO(s)  1 H2O(l) + 1 Sr(NO3)2(aq)
2 H+ + SrO(s)  H2O +
19. a. Oxidizing agents cause (oxidation/reduction) in another
atom by the process of (oxidation/reduction), which
results in the agent (gaining/losing) electrons.
b. Reducing agents cause (oxidation/reduction) in another
atom by the process of (oxidation/reduction), which
results in the agent (gaining/losing) electrons.
20.
SO32O = -2; S + 3(-2) = -2  S = +4
Sr2+
MnO4
-
Cr2O7
O = -2; Mn + 4(-2) = -1  Mn = +7
only works if H = +1 and O = -1
H2O2
2-
O = -2; 2 Cr + 7(-2) = -2  Cr = +6
21.
2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
.30 L x .40 mol/L = .12 mol Ba2+ – (.20 L x .20 mol/L =
.040 mol CO32-) = 0.080 mol Ba2+/0.50 L = 0.16 M
.1 L x 1 mol/L = .1 mol PO43- + .3 mol Na+
.1 L x 1 mol/L = .1 mol Ag+ + .1 mol NO3- (Ag+ = 0)
13.
B
Pb2+ is insoluble in Cl-, All ions are insoluble in S2- and
OH-, Cu2+ forms complex with NH3, others form OH- ppt.
14.
C
H+ + OH-  H2O
The precipitate is Pb3(PO4)2(s). Only Pb2+ and PO43- are
reactants  3 Pb2+ + 2 PO43-  Pb3(PO4)2(s)
11.
hydrochloric acid
+
Metal (cation) plus nonmetal (anion) does not use
prefixes. The ending for a non-polyatomic ion is ide
10.
B
OH-
Copper (II) is Cu2+ and sulfate is SO42-  CuSO4
9.
perchloric acid
1 H2SO4(aq) + 2 KOH(aq)  2 H2O(l) + 1 K2SO4(aq)
perchlorate = ClO4-, chlorate = ClO3-, chlorite = ClO2and hypochorite = ClO-
8.
sulfurous acid
18.
.025 L x .12 mol Ba... x 2 mol HCl x
1L _ = 0.040 L
1L
1 mol Ba... .15 mol HCl
7.
sulfuric acid
H+
.32 L NaOH x .5 mol/L x 1 mol HC.../1 mol = .16 mol HC...
0.16 mol HC.../0.020 L = 0.80 M
All NH4+, K+ and NO3- salts are soluble. Most
carbonates are insoluble including BaCO3.
15.
C
KMnO4 is a common oxidizing agent that is purple and
turns clear when reduced. (Used in the % peroxide lab)
16.
B
CuSO4 is light blue. Cu2+ compounds vary from yellowblue to green-blue
17.
A
18.
D
PbSO4 and KCl are white, but KCl is soluble whereas
PbSO4 is insoluble.
Ni2+ + 2 OH-  Ni(OH)2(s), Ag+ + Cl-  AgCl(s),
Ba2+ + SO42-  BaSO4(s)  only Na+ is spectator.
19.
C
20.
Cu2+ + 4 NH3  Cu(NH3)42+
light blue
dark blue
B
Pb2+ is soluble in hot and forms a precipitate with
CrO42- (PbCrO4)  Pb2+. Ag+ is soluble in NH3
The phenolphthalein would turn pink because of the
presence of OH- in solution making it basic.
b. (1)
21.
C
Without hydro  oxyacid, "ous" ending "ite" anion
= H2SO3
Ba2+ + 2 F-  BaF2(s)
22.
D
(2)
KF is the limiting reactant because it takes two moles
of KF to react with 1 mol of Ba(NO3)2  half of the
Ba(NO3)2 will remain after all of the KF has reacted.
c. (1)
2H2SO4: 2+S–8=0 (S=6), H2SO3: 2+S–
2O3
: 2S–6=-2 (S=4), S2-: S=-2, SO2Cl2: S–4 –2=0 (S=6)
23.
C
24.
B
ox.# 1 1 -2
0
1 7 -2
2 HClO + 3 O2  2 HClO4
Sr(OH)2 + 2 H+  2 H2O + Sr2+
Zn + 2 H+  Zn2+ H2(g)
(CuS  H2S, CaCO3  CO2, Mg(OH)2  H2O)
(2)
0.500 L H+ x 0.40 mol H+/1 L x 1 mol OH-/1 mol H+ x 1
mol Sr(OH)2/1 mol OH- = 0.100 mol Sr(OH)2
d. (1)
25.
B
26.
A
ox # 0
1
5 -2
2
2 -2
1 -2
3 Cu + 8 H+ + 2 NO3-  3 Cu2+ + 2 NO + 4 H2O
2.5 mol I2 x 10 mol HI = 5 mol HI
5 mol I2
27.
A
0.20 L x 0.20 mol MnO4- x 3 mol ClO2- = 0.030 mol
1L
4 mol MnO4-
28.
C
0.014 L x 0.10 mol MnO4- x 5 mol Fe3+ = 0.0070 mol Fe3+
1L
1 mol MnO40.025 L
29.
B
The solid product is made from ions.
30.
D
The product contains anions attached to a metal
cation, but the product has an overall charge.
31.
A
The compound MgO is made from its elements when
Mg is oxidized and O is reduced.
32.
C
The solid product is made from ions.
33.
D
The product contains anions attached to a metal
cation, but the product has an overall charge.
34.
B
The bromine (ox # = 0) becomes Br- (ox # = -1) and
BrO3- (ox # = +5).
35.
A
Combustion reactions are also redox reactions, where
the oxidizing agent is oxygen gas.
36.
D
K is oxidized and Br is reduced.
37.
B
The solid product is made from ions.
38.
C
The product contains anions attached to a metal
cation, but the product has an overall charge.
39.
A
H+ and OH- combine to form H2O (counterions form the
aqueous salt).
Practice Free Response
1.
a.
(1)
Li2O + H2O  2 Li2+ + 2 OH(2)
Ag+ + Fe2+  Ag + Fe3+
(2)
The precipitated solid silver remains on the filter
paper.
2. a.
0.08843 L OH- x 0.102 mol OH- x 1 mol HA = 0.00902 mol
1L
1 mol OH1.625 g/0.00902 mol = 180. g/mol
b.
It will take more drops of NaOH to reach equivalence 
the calculated moles of acid would be to be too high and
the molar mass to be too low.
3. a.
0.0135 L x 0.0200 mol MnO4- = 2.70 x 10-4 mol MnO41 L MnO4b.
2.70 x 10-4 mol MnO4- x 5 mol Fe2+ = 1.35 x 10-3 mol Fe2+
1 mol MnO4c.
1.35 x 10-3 mol Fe2+ x 55.8 g Fe2+ = 0.0753 g Fe2+
1 mol Fe2+
d.
0.0753 g Fe2+/0.500 g x 100 = 15.1% Fe2+
4.
a.
MM = 9.01 + 2(12.0) + 4(16.0) + 3(18.0) = 151 g/mol
% C = [2(12.0)/151] x 100 = 15.9 %
b. (1)
3.21 g BeC2H4•3 H2O x 97.0 g BeC2O4 = 2.06 g BeC2O4
151 g BeC2O4•3 H2O
(2)
3.21 g BeC2H4•3 H2O x 1 mol x 3 mol H2O = 0.0638 mol
151 g 1 mol Be...
PV = nRT
V = nRT = (0.0638 mol)(0.0821 atm•L/mol•K)(493 K)
P
(735/760 atm)
V = 2.67 L
c. (1)
C2O42- is the reducing agent (the oxidation number for
C in C2O42- is +3 and in CO2 is +4, which is an increase
 C2O42- is acting as an reducing agent)
(2) MnO40.01780 L MnO4- 0.0150 mol = 2.67 x 10-4 mol MnO41L
C2O422.67 x 10-4 mol MnO4- x 5 mol C2O42- = 6.68 x 10-4 mol C2O422 mol MnO4(3)
100 mL x 6.68 x 10-4 mol C2O42- = 3.34 x 10-3 mol C2O4220 mL
(4)
.00334 mol C2O42- x 1 mol BeC2O4 x 97.0 g =.324 g BeC2O4
1 mol C2O42- 1 mol Be...
0.324 g/0.345 g x 100 = 93.9 %