CHAPTER 3 HOMEWORK ANSWERS
1)3.2
a) 1 mol C12H22O11 contains 12 mol C
b) C atoms = (12 mol C)(6.022 x 1023 C atoms/mol C)
= 7.226 x 1024 C atoms
2)3.9
a)
b)
c)
d)
(NH4)3PO4
CH2Cl2
CuSO4·5H2O
BrF5
3)3.12
a)
m (g) KMnO 4 = 0.57 mol KMnO 4 x
3(14.01)
12.01 +
63.55 +
79.90 +
+ 12(1.008) + 30.97 + 4(16.00)
2(1.008) + 2(35.45)
32.07 + 9(16.00) + 10(1.008)
5(19.00)
=
=
=
=
149.10 g/mol
84.93 g/mol
249.72 g/mol
174.90 g/mol
158.04 g KMnO 4
1 mol KMnO 4
= 9.0 x 101 g KMnO4
b)
1 mol Mg(NO3 ) 2
6 mol O atoms
x
148.33 g
1 mol Mg(NO3 ) 2
O atoms (mol) = 8.18 g Mg(NO3 )2 x
= 0.331 mol O atoms
c)
O atoms = 8.1 x 10−3 g CuSO4 ⋅ 5H 2 O
x
1 mol CuSO4 ⋅ 5H 2 O
9 mol O atoms
6.022 x 1023 O atoms
x
x
249.7 g
1 mol
1 mol O atoms
= 1.8 x 1020 O atoms
4)3.14
151.01 g
= 97 g MnSO 4
1 mol
a)
m (g) = 0.64 mol MnSO 4 x
b)
n (mol) Fe(ClO 4 )3 = 15.8 g Fe(ClO4 )3 x
1 mol
= 4.46 x 10−2 mol Fe(ClO4 )3
354.2 g
c)
N atoms = 92.6 g NH 4 NO2 x
1 mol
2 mol N
6.022 x 1023 N atoms
x
x
64.05 g
1 mol NH 4 NO3
1 mol N
= 1.74 x 1024 N atoms
5)3.17
a)
m (g) = 3.52 mol Cr2 (SO 4 )3 ⋅ 10H 2 O x
b)
572.4 g
= 2.01 x 103 g Cr2 (SO 4 )3 ⋅ 10H 2 O
1 mol
m (g) = 9.64 x 10 24 molecules Cl 2 O 7 x
= 2.93 x 103 g Cl2 O7
1 mol Cl2 O7
6.022 x 10
23
molecules
x
182.9 g
1 mol
c)
mol Li 2SO 4 = 56.2 g Li 2SO 4 x
1 mol
= 0.511 mol Li 2SO 4
109.95 g
FU Li 2SO 4 = 0.511 mol Li 2SO 4 x
6.022 x 1023 formula units
1 mol
= 3.08 x 10 23 formula units Li 2 SO 4
d)
ions Li + = 0.511 mol Li 2SO4 x
2 mol Li +
6.022 x 1023 Li +
x
1 mol Li 2SO4
1 mol Li +
ions SO 4 2− = 0.511 mol Li 2SO 4 x
= 6.15 x 1023 Li + ions
1 mol SO 4 2 −
6.022 x 1023 SO 4 2−
x
1 mol Li 2SO 4
1 mol SO 4 2−
= 3.08 x 10 23 SO 4 2 − ions
atoms S = 0.511 mol Li 2SO4 x
atoms O = 0.511 mol Li 2SO 4 x
6)3.20
a)
2 C
CsC 2 H 3O 2
=
1 mol S atoms
x 6.022 x 1023
1 mol Li 2SO 4
4 mol O atoms
x 6.022 x 1023
1 mol Li 2SO 4
= 3.08 x 1023 S atoms
= 1.23 x 1024 O atoms
2 (12.01)
132.9 + 2 (12.01) + 3 (1.008) + 2 (16.00)
=
0.1251 mass
fraction C
b)
9 (16.00)
9O
= 0.3428 mass fraction
=
238.03 + 32.07 + 6 (16.00) + 3 (18.02)
UO 2 SO 4 ⋅ 3H 2 O
O
7)3.22
atoms O = 38.0 g O 2 x
1 mol O2
2 mol O atoms
6.022 x 1023 atoms
x
x
32.00 g O 2
1 mol O 2
1 mol O atoms
= 1.43 x 10 24 O atoms
8)3.29
g Fe 2 +
g Hb ⎞
⎛
⎜ 6.8 x 10 4
⎟ (0.33 % Fe 2 + ) = 2.24 x 10 2
1 mol Hb
mol ⎠
⎝
2.24 x 10 2 g Fe 2 +
1 mol Fe 2+
4.02 mol Fe 2 +
x
=
1 mol Hb
1 mol Hb
55.85 g Fe 2 +
≡ 4 Fe 2+ ions/molecule Hb
9)3.33
a)
C2H4 = M.F.
∴ CH2 = E.F.
E.F. mass = 12.01 + 2.016 = 14.03
b) C2H6O2 = M.F.
CH3O = E.F.
E.F. mass = 12.01 + (3)(1.008) + 16.00 = 31.03
c)
N2O5 = M.F. = E.F.
(2)(14.01) + (5)(16.00) = 108.02
d) Ba3(PO4)2 = M.F. = E.F.
3(137.3) + 2(30.97) + 8(16.00) = 601.8
e)
Te4I16 = M.F.
TeI4 = E.F.
E.F. mass = 127.6 + 4(126.9) = 635.2
0.063 mol Cl
1
2
=
=
0.22 mol O
3.5
7
Cl2O7
b)
2.45 g Si
28.09 g/mol Si
0.0872
1
=
=
12.4 g Cl
0.3498
4
35.45 g/mol Cl
SiCl4
c)
27.3 % C
12.01 g/mol C
2.28
1
=
=
72.7 % O
4.54
2
16.00 g/mol O
10)3.37 a)
CO2
11)3.41 M + F2 → MF2
a)
0.600 mol of a metal M, therefore,
(2)(0.600) = 1.20 mol F
b) 1.20 mol F = (1.20)(19.00) = 22.8 g F
46.8 g MF2 − 22.8 g F = 24.0 g M
c)
12)3.51 a)
b)
c)
d)
13)3.52 a)
b)
c)
d)
14)3.53 a)
b)
c)
d)
M has
24. g
= 40 amu molar mass, so M is calcium .
0.60 mol
16Cu(s) + S8(s) → 8Cu2S(s)
P4O10(s) + 6H2O(l) → 4H3PO4(l)
B2O3(s) + 6NaOH(aq) → 2Na3BO3(aq) + 3H2O(l)
4CH3NH2(g) + 9O2(g) → 4CO2(g) + 10H2O(g) + 2N2(g)
Cu(NO3)2(aq) + 2KOH(aq) → Cu(OH)2(s) + 2KNO3(aq)
BCl3(g) + 3H2O(l) → H3BO3(s) + 3HCl(g)
CaSiO3(s) + 6HF(g) → SiF4(g) + CaF2(s) + 3H2O(l)
(CN)2(g) + 4 H2O(1) → H2C2O4(aq) + 2 NH3(g)
2 SO2(g) + O2(g) → 2SO3(g)
Sc2O3(s) + 3H2O(l) → 2Sc(OH)3(s)
H3PO4(aq) + 2NaOH(aq) → Na2HPO4(aq) + 2H2O(l)
C6H10O5(s) + 6O2(g) → 6CO2(g) + 5H2O(g)
15)3.54 a)
b)
c)
d)
As4S6(s) + 9O2(g) → As4O6(s) + 6SO2(g)
2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) → P4(g) + 6CaSiO3(l) + 10CO(g)
3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)
6S2Cl2(l) + 16 NH3(g) → S4N4(s) + S8(s) + 12 NH4Cl(s)
16)3.61 4HCl(aq) + MnO2(s) → MnCl2(aq) + 2H2O(g) + Cl2(g)
a)
n (mol) Cl 2
b)
m (g) Cl2
1 mol Cl2
= 0.455 mol Cl 2
4 mol HCl
= 1.82 mol HCl x
= 0.455 mol Cl2 x
70.90 g Cl2
= 32.3 g Cl2
1 mol Cl2
17)3.62 Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g)
a)
n (mol) Bi 2 O3 = 352 g Bi 2 O3
b)
n (mol) Bi = 0.755 mol Bi 2 O3 x
x
1 mol Bi 2 O3
= 0.755 mol Bi 2 O3
466.0 g Bi 2 O3
2 mol Bi
= 1.51 mol Bi
1 mol Bi 2 O3
18)3.67 P4(s) + 10Cl2(g) → 4PCl5(s)
m (g) Cl2 = 355 g P4 x
1 mol P4
10 mol Cl2
x
123.9 g
1 mol P4
x
70.90 g Cl2
= 2.03 x 103 g Cl2
1 mol Cl2
19)3.68 S8(s) + 24F2(g) → 8SF6(s)
m (g) F2 = 17.8 g S8 x
1 mol S8
24 mol F2
38.00 g F2
x
x
256.56 g S8
1 mol S8
1 mol F2
= 63.3 g F2
20)3.71 2Ca(s) + O2(g) → 2CaO(s)
a)
n (mol) CaO = 4.20 g Ca x
2 mol CaO
1 mol Ca
= 0.105 mol CaO
x
2 mol Ca
40.08 g
b)
n (mol) CaO = 2.80 g O 2 x
1 mol O 2
2 mol CaO
x
= 0.175 mol CaO
32.00 g
1 mol O 2
c)
The Ca is the limiting reagent.
d) Theoretical yield is therefore based on limiting reagent.
m (g) CaO = 0.105 mol CaO x
56.08 g CaO
= 5.89 g CaO
1 mol CaO
21)3.72 SrH2(s) + 2H2O(l) → Sr(OH)2(s) + 2H2(g)
a)
n (mol) H 2 = 5.63 g SrH 2 x
b)
n (mol) H 2 = 4.80 g H 2 O x
c)
SrH2 is the limiting reagent.
d)
22)3.76
2H2(g)
↑
1 mol SrH 2
2 mol H 2
x
= 0.126 mol H 2
89.64 g SrH 2
1 mol SrH 2
2 mol H 2
1 mol H 2 O
= 0.266 mol H 2
x
2 mol H 2 O
18.02 g
m (g) H 2 = 0.126 mol H 2 x
+
0.0359 g ≡ 0.0178 mol
2.016 g H 2
= 0.253 g H 2
1 mol H 2
O2(g) → 2H2O(l)
↑
0.0175 mol
n (mol) O 2 = 0.0178 mol H 2 x
1 mol O 2
= 0.00889 mol O 2 needed
2 mol H 2
∴ O2 is in excess by (0.0175 − 0.00889) = 0.0086 mol
m (g) O 2 = 0.0086 mol O 2 x
32.00 g O 2
= 0.28 g O 2 in excess
1 mol O 2
n (mol) H 2 O = 0.0178 mol H 2 x
2 mol H 2 O
= 0.0178 mol H 2 O
2 mol H 2
m (g) H 2 O = 0.0178 mol H 2 O x
18.02 g
= 0.321 g H 2 O formed
1 mol
23)3.82 PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(g)
m (g) HCl = 200. g PCl3 x
1 mol PCl3
3 mol HCl
36.46 g HCl
x
x
137.3 g
1 mol PCl3
1 mol HCl
=
159 g HCl
128 g HCl (actual)
x 100 % = 80.5 % yield
159 g HCl (theoretical)
24)3.83 CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
n (mol) CH 4 = 18.5 g CH 4 x
n (mol) Cl 2 = 43.0 g Cl 2 x
1 mol CH 4
= 1.15 mol CH 4
16.04 g CH 4
1 mol Cl 2
= 0.606 mol Cl 2
70.90 g Cl 2
Cl2 is the limiting reactant.
m (g) CH3Cl = 0.606 mol Cl2 x
CH3Cl
1 mol CH3Cl
50.48 g CH3Cl
x
x .800
1 mol Cl2
1 mol CH3Cl
=
24.5 g
25)3.92 a)
C x V = amount of solute
m (g) = 0.207 M x 0.1758 L x
b)
c)
26)3.94 a)
158.17 g Ca(CH3COO)2
= 5.76 g Ca(CH3COO)2
1 mol Ca(CH3COO) 2
⎛ 1 mol KI ⎞
(21.1 g KI) ⎜
⎟
⎝ 166.0 g ⎠ = 0.254 M
Molarity KI =
0.500 L
n (mol) NaCN =
0.850 mol NaCN
x 145.6 L = 124 mol NaCN
1 L of solution
m (g) K 2 SO 4 = 0.475 L x
5.62 x 10 −2 mol K 2 SO 4
174.27 g K 2 SO 4
x
1L
1 mol K 2 SO 4
= 4.65 g K 2 SO 4
b)
Molarity CaCl2 =
c)
ions Mg 2+ =
x
27)3.96 a)
2+
V1M1 = V2M2
M2 =
1.00 x 10
−3
L
= 5.90 x 10−2 M CaCl2
0.184 mol MgBr2
1 mol Mg 2+
x 1.00 x 10 −3 L x
1L
1 mol MgBr2
6.022 x 1023 Mg 2 +
1 mol Mg
6.55 x 10−3 g CaCl2
110.98 g/mol
= 1.11 x 1020 Mg 2 + ions
(37.00 mL)(0.250 M) = (150.00 mL)(M2)
(37.00 mL) (0.250 M)
= 0.0617 M
150.00 mL
b) (25.71 mL)(0.0706 M) = (500.00 mL) (M2)
M2 =
c)
(25.71 mL) (0.0706 M)
= 0.00363 M
500.00 mL
mol Na+ from NaCl = 3.58 x 10-3 L NaCl x
0.288 mol NaCl
1 mol Na +
x
1 L NaCl
1 mol NaCl
= 1.03 x 10-3 mol Na+
mol Na+ from Na2SO4 =
0.500
+
6.51 x 10 −3 mol Na 2 SO 4
2 mol Na
x
1 L Na 2 SO 4
1 mol Na 2 SO 4
L
Na2SO4
= 6.51 x 10−3 mol Na+
molarity of Na+ =
(1.03 x 10 −3 + 6.51 x 10 −3 ) mol Na +
= 0.0150 M Na+
0.504 L solution
x
28)3.98 a)
b)
m (g) = 1000 mL soln x
Concentration (M) =
0.700 g HNO 3
1.41 g soln
x
= 987 g HNO 3
1 mL soln
1 g soln
987 g HNO3
1 mol HNO3
x
= 15.7 M HNO3
1L
63.02 g HNO3
2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)
29)3.100
Volume (mL) = 16.2 g CaCO3 x
1 mol CaCO3
2 mol HCl
1L
103 mL
x
x
x
100.09 g CaCO3
1 mol CaCO3
0.383 mol HCl
1 L
= 8.45 x 102 mL HCl
30)13.58a)
C(m) NH2CH2COOH =
b) C(m)C3H8O3 =
88.4 g NH 2 CH 2 COOH x
1.250 kg H 2 O
1 mol
75.07 g
= 0.942 m
8.89 g C3 H8 O3 x (1 mol C3 H8 O3 / 92.09 g C3 H8 O3 )
= 1.29 m
0.0750 kg C6 H 6
1 mol
36.46 g
= 5.97 m
0.753 kg H 2 O
164 g HCl x
31)13.59a)
C(m)HCl =
b) C(m)C10H8 =
32)13.64a)
b)
c)
33)13.68
16.5 g C10 H8 x (1 mol C10 H8 /128.2 g C10 H8 )
= 2.41 m
0.0533 kg C6 H 6
0.30 mol isopropanol
= 0.27
0.30 mol isopropanol + 0.80 mol H 2 O
( 0.30 mol )( 60.10 g / mol )
( 0.30 ) ( 60.10 ) + ( 0.80 )(18.02 )
x 100 = 56% w/w
0.30 mol isopropanol
= 21 m
(0.80)(18.02)
kg H 2 O
1000
100 g soln − 8.00 g NH3 = 92.0 g H2O
⎛ 1 mol NH3 ⎞
(8.00 g NH3 ) ⎜
⎟
17.03 g NH3 ⎠
0.470 mol NH3
⎝
=
= 5.11 m
C(m)NH3 =
0.0920 kg H 2 O
⎛ 1 kg H 2 O ⎞
(92.0 g H 2 O) ⎜ 3
⎜ 10 g H O ⎟⎟
2 ⎠
⎝
C(M) NH3 =
1 mol NH3
8.00 g NH 3
103 mL
0.9651 g soln
x
x
x
17.03 g NH3
100 g soln
1 mL soln
1L
= 4.53 M
XNH3 =
0.470 mol NH3
= 0.0842 mole fraction
0.470 + 5.11 mol H 2 O
34)4.14 a)
0.25 mol NH4+ + 0.25 mol Cl− = 0.50 mol total.
b) n(mol)
=
26.4 g Ba(OH)2⋅8H2O
1 mol Ba(OH)2 ⋅ 8H 2 O
315.4 g Ba(OH)2 ⋅ 8H 2 O
x
=
0.0837 mol
Ba(OH)2⋅8H2O
So 0.0837 mol Ba2+ + 0.167 mol OH− = 0.251 mol total.
c)
1 mol
n (mol) = 1.78 x 10 20 FU x
6.022 x 10
23
= 2.96 x 10−4 mol LiCl
FU
So 2.96 x 10-4 mol each of Li+ and Cl-, total of 5.91 x 10-4 mol.
35)4.18 a)
n (mol) Al3+ =
n (mol) Cl− =
2.45 mol AlCl3
103 mL
2.45 mol AlCl3
x 95.5 mL x
x 95.5 mL x
103 mL
3 mol Cl−
= 0.702 mol Cl−
1 mol AlCl3
6.022 x 1023 Al3+ ions
ions Al3+ = 0.234 mol Al3+ x
1 mol Al3+ ions
6.022 x 1023 Cl− ions
ions Cl− = 0.702 mol Cl− x
b)
1 mol Al3+
= 0.234 mol Al3+
1 mol AlCl3
1 mol Cl− ions
= 1.41 x 1023 Al3+ ions
= 4.23 x 1023 Cl− ions
4.59 g Na 2SO 4
1 mol Na 2SO 4
2 mol Na +
x
x
1 L soln
142.05g Na 2SO 4
1 mol Na 2SO 4
n (mol) Na + = 2.50 L soln x
= 0.162 mol Na+
n (mol) SO42− = 2.50 L soln x
2 mol SO42−
4.59 g Na 2SO4
1 mol Na 2SO4
x
x
1 L soln
142.05g Na 2SO4
1 mol Na 2SO4
= 0.0808 mol SO42−
ions Na + = 0.162 mol Na + x
6.022 x 1023 Na +
1 mol Na +
ions SO 42− = 0.0808 mol SO 42 − x
c)
FU MgBr2 = 80.5 mL x
2.16 x 1021 FU Mg Br2 x
1L
10
3
mL
= 9.73 x 1022 Na + ions
6.022 x 1023 SO 42−
1 mol SO 42−
x
= 4.86 x 1022 SO 42 − ions
2.68 x 10 22 FU
= 2.16 x 1021 FU MgBr2
1L
1 Mg 2+
= 2.16 x 1021 Mg 2+ ions
1 FU MgBr2
2.16 x 1021 Mg2+ ions ÷ 6.022 x 1023 = 3.58 x 10–3 mol Mg2+
2.16 x 10 21 FU MgBr2 x
2 Br −
= 4.31 x 10 21 Br − ions
1 FU MgBr2
4.32 x 1021 Br ÷ 6.022 x 1023 = 7.17 x 10–3 mol Br–
36)4.20 a)
n (mol) = 0.140 L x
b)
n (mol) = 6.8 mL x
2.5 mol HClO 4
1 L
1L
3
10 mol
x
= 0.35 mol HClO4 → 0.35 mol H+(aq)
0.52 mol HNO 3
1L
= 0.0035 mol HNO3 → 0.0035
mol H+(aq)
c)
n (mol) = 2.5 L x
0.056 mol HCl
= 0.14 mol HCl → 0.14 mol H+(aq)
1L