Page 342 - Theory and Problems of BEGINNING CHEMISTRY

PRACTICE QUIZZES 331

Chapter 7

1 mol Co 1 mol O
1. 71.05 g Co = 1.21 mol Co 28.95 g O = 1.81 mol O
58.9gCo 16.0gO
1.81 mol O 1.50 mol O 3 mol O
= =
1.21 mol Co 1.00 mol Co 2 mol Co
The formula is Co 2 O 3 .

1 mol CaCO 3 3 mol O
2. 175 g CaCO 3 = 5.25 mol O
100 g CaCO 3 1 mol CaCO 3

Chapter 8
1. (a)2 C 3 H 6 O 2 + 7O 2 −→ 6CO 2 + 6H 2 O
(b)4 Al + 3O 2 −→ 2Al 2 O 3
2. (a)Zn + Cu(NO 3 ) 2 −→ Cu + Zn(NO 3 ) 2
(b) BaO + SO 3 −→ BaSO 4

Chapter 9
1. (a)H + OH −→ H 2 O
−
+
− − (The acid is weak.)
(b) HClO 2 + OH −→ H 2 O + ClO 2
(c)2 H + Ba(OH) 2 (s) −→ Ba 2+ + 2H 2 O [The Ba(OH) 2 is not in solution.]
+
2. 4Ce 4+ + 2 CuI −→ I 2 + 4Ce 3+ + 2Cu 2+
(First balance the iodine, then the copper. The cerium has to provide for four positive charges, so there are four cerium
ions on each side.)

Chapter 10
1. 2 NaI + Pb(NO 3 ) 2 −→ PbI 2 + 2 NaNO 3
1 mol NaI 1 mol PbI 2 461 g PbI

105 g NaI 2 = 161 g PbI 2
150 g NaI 2 mol NaI 1 mol PbI 2

1 mol CaCl 2
2. 24.0 g CaCl
2 = 0.216 mol CaCl 2
111 g CaCl 2

1 mol AgNO
52.0 g AgNO 3 = 0.306 mol AgNO
3 3
170 g AgNO 3
2 AgNO + CaCl 2 −→ Ca(NO 3 ) 2 + 2 AgCl
3
0.306 mol AgNO
3 = 0.153 mol AgNO
2 3
0.216 mol CaCl 2
= 0.216 mol CaCl 2
1
AgNO 3 is in limiting quantity.
All in moles: 2 AgNO 3 + CaCl 2 −→ Ca(NO 3 ) 2 + 2 AgCl
Initial: 0.306 0.216 0.000 0.000
Change: 0.306 0.153 0.153 0.306
Final: 0.000 0.063 0.153 0.306