CHAPTER 6
Chemical Reactions: An Introduction
CHAPTER ANSWERS
1.
The types of evidence for a chemical reaction mentioned in the text are a change in color,
formation of a solid, evolution of a gas, and absorption or evolution of heat. Other bits of evidence
that might also be observed include appearance or disappearance of a characteristic odor or
separation of the reaction mixture into layers of visibly different composition.
2.
Most of these products contain a peroxide, which decomposes releasing oxygen gas.
3.
The fact that the material in the drain that did not dissolve in water dissolves when the drain
cleaner is added suggests that rather than simple dissolving, the material in the drain has
undergone a chemical change that makes it soluble. You may also have noticed that the drain
cleaner evolved heat when added to the drain; evolution or absorption of heat is also often a sign
of a chemical reaction.
4.
Bubbling takes place as the hydrogen peroxide chemically decomposes into water and oxygen
gas.
5.
The container of a flashlight battery usually consists of zinc, which is one of the substances
involved in the chemical reaction in the battery that generates the electricity. The fact that the
zinc decays until the battery leaks is a sign that a chemical reaction has taken place.
6.
The two components are both liquids, but harden to a solid when combined. There is also heat
evolved during the reaction.
7.
A and B are the reactants; C and D are the products; the arrow indicates that a reaction takes
place.
8.
atoms
9.
the same as
10.
the same
11.
The physical state is included because it helps us visualize the reaction taking place (for example,
if a solid forms when two liquids are mixed together, that is very significant). The physical state is
also included because it may influence some measured properties of the reaction. For example, if
water vapor is formed in a reaction, a different quantity of energy may be involved than if liquid
water were formed (See Chapter 10.). We indicate the physical states as follows: (s), solid; (1),
liquid; (g), gas; (aq) aqueous solution.
12.
water
13.
Zn(s) + HCI(aq) —> 2
ZnC1
(
aq) + 2
H
(
g)
14.
H
(
0
2
aq)
15.
H(g) + 02(g)
—
H
(
2
g) +
—
02(g)
H
0
2
(g)
Copyright © Hougtiton Mifflin Company. All rights reserved.
86
Chapter 6: Chemical Reactions: An Introduct
ion
16.
3
Ag
(
aq
NO
)
HC1(aq)
+
3
Pb
(
2
)
aq
(N)
0
[ICI(aq)
+
AgCJ(s)
—
Ag(s)
18.
C3HX(g)
+
02(g)
—*
2
C
(
g)
0
C
(
H
3
g)
+
0
(
2
g)
-
C0(g)
+
19.
B
(
3
0
2
s) + Mg(s)
—
B(g)
MgO(s)
20.
3
Ca
(
s)
CO
21.
P
(
4
s)
22.
2
Si
(
s)
0
23.
3
N
N
4
(
s
H
0
)
24.
11
S
2
(g) +
25.
C
(
H
2
g) + 02(g)
26.
S0
(
2
g)
27.
+
+
-
N
0
2
(g)
+
+
+
2
C
(
g)
0
-
2
C
(
g)
0
3
S
2
H
(
aq
0
)
H
0
2
(1)
—
4
S
2
H
(
aq
0
)
—*
Ba(s)
+
2
A
(
3
0
s)
L
CaO(s)
+
Al(s)
—
Ca(s)
+
2
A
(
3
0
s)
1
SrO(s)
F
Al(s)
—
Sr(s)
28,
N0(g)
+
Oi(g)
-
2
N
(
g)
0 +02(g)
29.
4
CF
(
g)
I
+
C1
(
2
g)
4
CC
(
l)
J + HCI(g)
30.
NH
(
3
g)
+
3
HN
(
aq
O
)
-
31.
CaO(s)
+
112 0(g)
32.
Xe(g)
F
(
2
g)
33.
3
NH
(
s
N0
)
34.
NHC1(s)
—*
+
‘:.o :(g.
:;
110(g)
4
Al(s)
36.
+
H
0
2
(g)
+
35.
[120(1)
C0(g)
—*
+
1-12(g)
H
0
2
(g)
[120(1)
—*
+
1-120(g)
+
S0
(
2
g)
O(g) —,
-->
N0(g)
+
BaO(s)
+
3
HN
(
aq
O
)
1120(g)
-
2
Ca
(
aq
CI
)
—
Si(s)
—*
-
+
*
3
PC
(
s)
I
-
C(s)
+
3
Ag
(
aq
NO
)
—
HCI(aq)
C1
(
2
g)
+
S0
(
3
g)
3
FI
(
aq
NO
)
3
HN
(
aq
O
)
2
Pb
(
s)
CI
—>
17.
+
+
.,
2
A
(
3
0
s)
1
3
N
N
4
(
s
11
0
)
Ca(0H)(s)
.
4
Xe
(
s)
F
N
(
2
g)
NaOH(s)
+
02(g) +1120(g)
heat
> 3
N
(
g)
H+H
0(g) 4 NaC1(s)
2
The subscripts in a formula really defin
e what compound is present since the subs
cripts represent
in what proportions the elements combine
to form the compound. Changing the subs
cripts would
be changing the identity of the compoun
d.
whole numbers
37.
a.
3I
FeC
4
KOH
-
3
Fe(
OH)
+
KC1
Balance chlorine: FeC
3
1 + KOH —* Fe(
3
OH) + 3KC1
Balance potassium: 3
FeCI + 3KOH —* 3
Fe(OFI) + 3KCI
Balanced equation: 3
Fe
(
aq
CI
) + 3KOH(aq) —* 3
Fe(
(
s
QH
)) +
3KC1(aq)
Copyright © Houghton Mifflin Company. All rights
reserved,
1
Charj{ar : Chemical Reactions: An Introduction
h.
3
1
Pb(C
)
2
0
{
KI
4
2 + KC
Pbi
0
3
H
2
Balance iodine: Pb(C
)
0
3
H
2
2K[
Balance potassium: 0
3
1
Pb(C
)
2
i
2
Phi
—b
2K1
+
Balanced equation: 2
Ph(C
(
)
0
3
H
aq)
c.
)
1
0
4
P
1
+
—
Balance hydrogen: 4
P
1
0
0
-F
0
2
H
4 6L12() —
-
f.
Sb + Cl
2
-I-
3
1
2KC
2
0
{
-1
—
Phl(s)
2 iO
KC
(aq)
2
4
P
3
4H
C)
C
-
—,
—
4
P
3
41{
(
aq)
0
LiOH
2Lk)H
—
Balanced equation: L
O(s) 1 HO(I)
2
2
Mn0
2
Ph[
0 — 4
2
H
P0
3
4H
-F
Balance lithium: I,i.
0 4 H
2
0
2
e.
3
1
KC
2
O
1
2K1(aq)
Balanced equation; P
(s)
1
0
4
0 + ói{
0(i)
2
0
2
Li
+
0 - 11
2
H
4
P
3
0
Balance phosphorus: P40J
0
d.
87
Mti
2LiOl-I(aq)
Fins equation is already balanced!
2
CO
+
—.
3
SbC1
This equation is more difficult to balance than it may appear. The problem arises in the fact
that there are two Cl atoms on the left side of the equation, whereas there are three Cl atoms
on the right side of the equation. To balance the chlorine atoms, we need to know the
smallest whole number intc which both two and. three divide. This number is six: we need to
adjust the coefficients of Cl
2 and ShCJ so that there will be six chlorine atoms on each side
of the equation.
Balance chlorine: Sb
2
3d
Balance antimony: 2Sh
+
4 +1120
CH
—
CO
3
2ShCI
2 - 2SbCi
3Ci
3
Balanced equation: 2Sb(s)
g.
—>
I
3Ci2(g)
-
CO ± 3112
Balaiced equation: 4
CH
(
g) ± H
0(g)
2
FeS
+
HCT
—
2ShCI
(
3
s)
H
1-
Balance hydrogen: CH
4+H
0
2
h.
—
—*
CO(g)
+
3H
(
2
g)
2+H
FeCI
S
2
Balance chlorine: FeS
-i-
2HCI
Balanced equation: FeS(s)
+
—*
2
FeC1
2HC1(aq)
-f
—*
S
2
H
FeC1
(
2
aq) -4 2
H
S
(g)
38.
a.
Al(s) + CuO(s)
—*
Al
(
3
0
2
s)
balance Al: 2A1(s) -I- CoO(s)
balance 0: 2A1(s)
balance Cu: 2A1(s)
-4
+
Cu(l)
--
—
A1
(
3
0
2
s) -I- Cu(/)
3CuO(s)
—+
AlO
(
3
s) + Cu(l)
3CuO(s)
—*
A1
(
3
0
2
s)
balanced equation: 2A1(s) + 3CuO(s)
Copyright © 1-loughton Miftlin Company M rights reserved.
---
+
3Cu(1)
A1
(
3
0
2
s) + 3Cu(l)
88
Chapter 6: Chemical React
ions An Introduction
b.
Sg(s)
+
F
(
2
g)
—
6
S
(
g
F
)
balance sulfur: 8
S
(
s) + F
(g)
2
8S
(
6
gF
)
balance fluorine: S
(s) + 24F(g)
8
8S
(
6
gF
)
balanced equation: S
(s) + 2
8
24
(
gF
)
8S
(
5
gF
)
c.
Xe(g) + F
(g) .—* 6
2
Xs
(
eF
)
balance fluorine: Xe(g) +
3g
(
2
F
) — 6
Xs
(
eF
)
balanced equation: Xe(g)
+ 2
3g
(
F
) — 6
Xs
(
eF
)
d.
N
C
4
I(H
g) + KOH(s) —* 3
N
(
g
H
)+H
0(g) + KC1(s)
2
The equation is already bal
anced.
e.
SiC(s) + 2
C
(
g1
) — 4
Si1
(
C)
1 + C(s)
balance chlorine: SiC(s) +
2C
(
2
g1
) — 4
SiC
(
I)
I + C(s)
balanced equation: SiCIs)
+ 2
2C
(
g1
) —* 4
S1
(
iC)
I + C(s)
f.
0(s) + H
2
K
0(1) — KO1{(aq)
2
balance potassium: 2
1
0
(s
(
) + 1120(1) — 2KOH(aq
)
balanced equation: 1(2
0(8) + 1120(l) —* 2K011
(aq)
g.
(g) + 1120(1) —* 3
5
0
2
N
H
(
a
N
q
O
)
balance nitrogen: N
(g) + 1120(1) —* 3
5
0
2
2H
(
aq
N)
0
balanced equation: N
(g) + 1120(1)
5
0
2
2H
(
3
aq
N)
0
h.
F
S
2
(gI
)+2
C
(
g1
)4 S
(s) + HCI(g)
8
balance sulfur: 2
8(g
S
H
)+2
C
(
g1
) 4 Sg(s) + HCI(g)
balance hydrogen: 2
8(g
S
H
) .f 2
8C
(
g1
)4S
(s) + J6HCI(g)
8
balanced equation: 8HS(g)
+ 2
8C
(
g1
)4S
(s) + 1611C1(g)
8
—
—
-
39.
a.
2 +1(1
Br
— 12+
KBr
Balance bromine: 2
Br + KI
Balance iodine: Br.,+ 2K1
—*
I
+
2KBr
- 12 + 2KBr
Balanced equation: (
2
B
gr
) + 2K1(aq)
(s)
2
1
+F1 2
K
0
2
-K0H+0
—
b.
i-
2KBr(aq)
Balance hydrogen: 1(2
02 + H
0 —> 21(011 + 0
2
Balance oxygen: 1(202
+ 1120
2KOH +3402
Convert to whole number
s: 21(202+21120 —* 41
(011+02
Balanced equation: 2
2s
(
0
1)
( 1-21120(l) —* 4KOH(aq
) + 02(g)
-
Copyright © Houghton Miff
lin Company. All rights reserve
d.
Chapter 6: Chemical Reactions: An Introduction
c.
LIOH
2
CO
3
C
2
Li
O
—*
Balance lithium: 2IiO}{
d.
H
(
2
)
t
2
CO
1
+
3
C
7
K
0
0
2
H
3
NNO
3
KNO
—
+
Balance potassium: 3
C0
2
K
COi(g)
+
3
HNO
+
Balance nitrate ion: K
3
C
2
0
e.
3
C
2
Li
O
—>
Balanced equation: 2LiOH(s)
+
3
2KN0
—
3
2F1N0
4
LiAIH
±
LiC1
3
All-I
+
i1
0
2
(s)
—
2
CO
+ [120 ±
3
2KN0
—,
2[1.N0
(
3
aq)
—
3
C
2
Li
(
s)
O
2
CO
-
3
Aid
0
2
H
+
—
Balanced equat.: (
3
C
2
K
s)
0
-f
89
1120
+
2
CO
2KN0
(
3
aq)
+
H
0
2
(1)
+
+
C0
(
2
g)
We have an interesting situation here. There are four hydrogen atoms on the left side of the
equation, but only three hydrogen atoms on the right side. The smalLest number that is
divisible by both thur and three is 12. So the simplest way to begin balancing this equation is
to take care of the hydrogen atoms so that there are 12 on each side.
Balance hydrogei. 3LiAIH
4
Balance lithium: 3LiAII{
4
3
Aid
F
3
Aid
+
Balanced equation: 4
3LiAIII
(
s)
This equation is already balanced.
g.
4
S
2
Na
O
C
S
2
Na
—
Balance carbon: Na
4
S
2
O
i-
C
—
2C
+
—b
Balanced equation: (
4
S
2
Na
O
s)
NaCI
+
4
S
2
H
0
—
4
S
2
Na
O
Balance sodium: 2NaC1
4
LiCI
3LiCI
+
4A1H
(
3
s)
-.-*
F
4
2
2C0
+
S
2
Na
+
—
2
2C0
Na
S
2
(s)
±
—
4
S
2
[1
0
4
4
S
2
Na
O
4
S
2
Na
O
—p
HSO(l)
—>
±
b.
3Fe(s)
c.
Ca(OH)
(
2
aq)
d.
Br
(
2
g)
e.
3NaOH(s)
f.
2NaNO
(
3
s)
g.
2Na
(
O
2
s)
h.
4Si(s•) + 8
S
(
s)
f
+
Caç’l(aq)
+
+
2HCI
4
S
2
Na
(
O
s)
H20(g)
4
4
+
FeO(s)
2HC1(aq)
2H
0
2
(I)
+
—
4
—
2NaNO
(
2
s)
21-120(1)
—
-4
1{2(g)
4
CaCI
(
2
aq)
—
SO(g)
4
P
3
H
(
aq)
0
—
CaSO
(
4
s) + 2NaCl(aq)
—
2C0
(
2
g)
HC1
40.
4
S
2
Na
(
aq)
O
3LiCI(s)
11(1
+
Balanced equation: 2NaC1(s)
S
2
Na
2C(s)
4
S
2
H
0
-
Balance chlorine: 2NaCI
a.
±
2
CO
+
Balance oxygen: Na
4
S
2
O
h.
3
4A111
—
A1CI
(
3
s)
+
f.
+
3
4A11{
—
+
2HBr(aq)
+
4
P
3
Na
(
aq)
O
+
—
21120(1)
4
S
2
H
(
aq)
0
3H
0
2
(l)
0
(
2
g)
4NaOH(aq)
2Si
(
4
S
2
s)
Copyright © Houghtor Muffin Company All rights reserved,
+
02(g)
+
2HCI(g)
90
Chapter 6: Chemical Reactions: An Intr
oduction
41.
a.
2Li(s)
1-
2
C
(
g)
1
—>
2LiC1(s)
b.
3 Ba(s)
-f
N
(
2
g)
—*
3
B
(
2
N
sa
)
c.
3
2N
(
aHQO
s
)
d.
2A1(s)
e.
2NiS(s)
3
N
C
2
(
s
O
a
)
—
6HCI(aq)
+
—*
+ 302(g) —*
2
C
(
g)
0
+
3
2A
(
aq
IC)
I
2NiO(s)
+
Ca[{(s)
g.
2112(g)
h.
2
2
(
3
0
s
B
)
+
6C(s)
a.
4NaCI(s)
+
2
2S
(
g0
)
b.
2
3B
(
l)
r
+
1
(
2
s)
c.
Ca(s)
2
2F
0
(g{
) —* (
2
Ca
a
(O
q
K)
)
d.
3
2B
(
gF
)
e.
2
S
(
g)
0
+
2
2C
(
g)
1
—>
2
SO
(
1
CI
)
f.
2
L
0
()
i
I-
H
0
2
(l)
--ì
2LiOH(aq)
g.
Mg(s)
CuO(s)
h.
3
F
(
4
0
se
)
a.
2
4K
(
s)
0
+
2
6E
0
([1
)
b.
2
F
(
3
0
se
)
+
3
6H
(
aq
N)
0
c.
3
4N
(
g
11
)
f
502(g)
d.
5
PC
(
l)
!
4H
0
2
(l)
e.
5
C
0
HH
(l) +302(g)
f.
2CaO(s)
g.
2
2M
(
s0S
) ± 702(g)
h.
3
Fe
(
s
CO
)+(
3
C
2
H
aq
0
)
a.
3
B
(
2
)
a
a(q
N)
0
b.
2
Pb
(
aq
C)
I
c.
C
O
5
H
2
H(l)
d.
2
Ca
(
s)
C+2
2H
0
(l)
Sr(s) -- 3
2H
(
aq
N)
0
21120(l)
C0(g)
+
—
H
0
2
(g)
3H
(
2
g)
2
2S
(
g0
)
f.
+
+
+
2
Ca
(
s
(OH)
)
+
.:
.:•
2H
(
2
g)
3
C
O
H
H
(l)
-
B
(
3
C
4
s)
—*
2
3C
(
g0
)
+
42.
+
+
—ì
-
4H
(
2
g)
F
0
(
2
g)
+
—+
4
2
S
2
(
N
s
0
)
a
+
4HCI(g)
21Br(s)
—
2
3
0
(g
H
)
+
21{20(g)
+
B
(
3
0
2
s)
+
+
12
1
(g)
6HF(g)
2
C
0
(g[
)
I
MgO(s)
+
Cu(1)
3Fe(l)
4-
2
4
0
(g
H
)
—*
43.
+
5C(s)
+
4KOH(aq)
—
—*
0(g)
4
2F
(
3
)
a
e(
q
N)
0
+
4N0(g)
—ì
—*
F
+
31120(1)
2
6
0
(g
H
)
4
j
P
3
(
aq
1
0
)
+
51{C1(g)
2C02(g)
+
3H
0
2
(l)
-+
4
(
2
0
aq
1{
)
2
2C
(
s
aC
)+2
C
(
g)
0
—>
—>
3
2M
(
so0
)
—,
4
4S0(g)
3
Fe
(
2
)
a
(H
q
CO
)
44.
e.
+
F
4
N
C
2
(
a
rO
q
a
)
4
S
2
K
(
aq
0
)
+
302(g)
-4
—>
BaCr0(s) + (
3
2N
aaN
qO
)
PbS0(s) -I- 2KC1(aq)
—>
2
2C
(
g0
) +31120(l)
2
Ca
(
s
(OH)
)+ 2
C
(
H
g)
— 3
Sa
(
2
)
r(q
N)
04 H
(g)
2
—
Copynght © Houghtori Mifflin Company
. PII rights reserved.
..•4•
k
Chapter 6: Chemical Reactions: An Introduction
f.
(s) -+- 4
2
Ba0
S
2
H
(
aq)
0
—
g.
(s)
3
2As1
(s)
2
31
h.
(aq)
4
2CuSO
45.
(g)
H
2
C
46.
Na(s)
2As(s)
—
+
+ 02(g) 4
4K1(s)
—*
(g)
2
C0
+
(s) -I- H0(1)
0
2
Na
(s) 4
3
KNO
48.
(aq)
C
2
H
1
0
2+H
1
0(1
2
49.
(g)
2
2H
50.
(s) +
3
0
2
2A1
51.
(s)
4
0
3
Fe
+
(g)
2
4F1
(s)
4
0
3
Fe
+
4C0(g)
2Li(s)
S(s)
52
+
2K(s)
+
2Rb(s)
53.
C
2
K
(
3
s)
0
C(s) —*
S(s)
—
+
(g)
2
4C0
—
3Fe(s)
+
0(g)
2
4H
3Fe(s)
—*
+
4C02(g)
+
(aq)
0
2
H
S(s)
2
Li
-
S(s)
2
Na
—
S(s)
2
K.
—>
2Cs(s) + S(s)
—
S(s)
2
Cs
2Fr(s)
S(s)
-
S(s)
2
Fr
Fe(s) -F 0(g)
—
FeO(s)
Fe(s) + 02(g)
—
(s)
3
0
2
Fe
54.
Ba0(s) + 1120(l)
55.
4B(s)
302(g)
5
1
2
4C
O
H(aq)
-I
(g)
2
3C0
S(s)
2
Rb
+
S
2
2K
(
4
aq)
0
C0(g) + N
(g)
2
+
+
—
+
+
+ 02(g)
4A1(s)
S(s)
+
(s)
2
1
OH(l)
3
CH
—
3C(s) —>
2Na(s) + S(s)
+
0(g)
2
H
NaOH(aq)
-
47.
C0(g)
2Cul(s)
+ H202(aq)
(s)
0
2
Na
+ 02(g) —4
+
+
(s)
4
BaSO
91
BaO(s)
—,
-4
(s) + 31120(l)
3
0
2
B
(s)
3
0
2
2B
-
(s)
3
2B(OH)
56.
(s)
3
2KC10
-
2KC1(s) + 3
O2(g)
57.
(aq)
0
2
2H
—*
2O(S)
211
58.
(g)
3
N}1
59.
(s) + 6HF(g)
3
CaSiO
60.
The senses we call “odor” and “taste” are really chemical reactions of the receptors in our body
with molecules in the food we are eating. Thp fact that the receptors no longer detect the “fishy”
odor or taste suggests that adding the lemon juice or vinegar has changed the nature of the amines
in the fish.
61.
Many over-the-counter antacids contain pither carbonate ion (C0
j or hydrogen carbonate ion
3
3 ). When either of these encounter stomach acid (primarily HCI), carbon dioxide gas is
(HCO
released.
62.
Fe(s) +
+
HC1(g)
S(s) -4
—>
+
02(g)
C1(s)
4
NH
-
(aq) + SiF
2
CaF
(g) + 3H
4
0(l)
2
FeS(s)
Copyright © Houghton Muffin Company. All rights reserved.
92
Chapter 6: Chemical Reactions: An
Introduction
63.
Na(s)
64.
4
C
2
K
(
arO
q)
65.
H
S
2
(g)
66.
2NaCI(aq)
+
2
2
0
(l
H
)
—
2NaBr(aq)
+
2
2
0
(l
H
)
—
2NaI(aq)
2
2
0
(l
H
)
2
C
(
g)
1
+
+
NaC1(s)
—.*
2
B
(
aq
aC1
)
2
Pb
(
a
(N
q
0j
)
)
+
+
—>
4
Ba
(
s
Cr)
O + 2KCI(aq)
PbS(s)
—*
+
2NaOH(aq) + 112(g)
2NaOH(aq) +112(g)
2NaOH(aq)
—
3
H
(
aq
NO
)
+
112(g)
+
2
C
(
g)
1
+
8r2(g)
+
12(g)
67.
Mg(s)
68.
2
C
(
s
aC
)
69.
P
(
4
s)
70.
CuO(s)
71.
PbS(s)
72.
3
N
S
2
(
aq
0
a
)
+
S(s)
a.
2
C
(
g)
1
+
2K13r(aq)
b.
4Cr(s) ÷ 302(g)
c.
P
(
4
s)
d.
2A1(s)
+
4
3
S
2
(
aH
q
0
)
—>
e.
3
PC
(
l)
1
+
2
3
0
(l
H
)
P
3
H
(
aq
0
)
f.
2S0(g) +02(g)
g.
C
(
1
H
7
l) +1102(g)
—>
h.
2
2
(
6
H
g
C
)
-*
a.
2
C
(
g)
1 + 2K1(aq)
b.
2
C
(
s
aC
)+0
2
2
(l
H
) —* 2
Cas
(
(OH)
)+ C
Lg)
H
2
2NaC1(s) + 4
S
2
H
(
l)
0 —* 4
N
S
2
(
s
O
a
) + 2HC1(g)
Cs)
(
2
aF + 4
S
2
H
(
l)
0 — 4
Ca
(
s
SO
) + 2HF(g)
02(g)
+
21120(l)
+
+ 02(g) —+
+
+
MgO(s)
—
C
(
H
2
g)
+
(g
1
0
4
P
0
)
4
S
2
R
(
aq
0
)
02(g)
2
Ca
(
s
(OH)
)
-
—*
4
C
(
a
uS
qO
)
PbO(s)
—*
-1- H0(l)
2
S
(
g)
0
+
N
(
3
0
2
S
aq
a
)
—
73.
+
6Ii(g)
+
—
—+
d.
f.
3BaO(s) +2A1(s)
g.
2A1(s) + 2
3F
(
g)
h.
2
C
(
g)
S + 3C12(g)
a.
SiCL(l) + 2Mg(s)
b.
2N0(g)
c.
2
3M
(
sn0
) + 4A1(s)
—*
4
A
(
2
(
3
)
a
S
q
1
0
)
+
+
3H
(
2
g)
3HC1(aq)
2S03(g)
—
—>
3
C
2
K
(
s0
)
2KCI(aq)
2
2
(
3
0
s
C)
r
—
702(g)
e.
+
3
4P
(
gH
)
—
74.
c.
2
B
(
1)
r
7CO(g) +81120(g)
4C02(g) + 6H0g
2KCI(aq) + 2
1
(
s)
K
0
2
(s)+ C02(g)
2
A
(
3
0
.)
1 .3Ba(s)
2A
(
3
s
1F
)
—,
-
CCL(l) + g
C1
2
S
)
—*
75.
+
2
C
(
g)
1
—*
—*
Si(s) + 2
2MgC
s
)
1
2NOC1(g)
—*
3Mn(s) + 0
2
2A
(
3
s)
1
Copyiight © Houghton Mifflin Com
pany. All rights reserved.
Chapter 6: Chemical Reactions: An Introduction
d.
i6Cr(s) + 8
3S
(
s)
e.
4NH
(
3
g)
f
AgS(s)
+
g.
302(g)
.—
h.
3
S
2
8Na
(
aq)
0
a.
Pb(N0
(
2
)
3
aq) + (
4
C
2
K
aq)
rO
—+
b.
BaCI
(
2
aq) + 4
S
2
Na
(
aq)
O
—*
BaSO(s)
c.
2CH
O
3
H(l)
2C0
(
2
g) +41120(g)
d.
3
C
2
Na
(
aq)
O + S(s) + 2
S0
(
g)
e.
Cu(s)
f.
Mu0(s)
+
4HCI(aq)
—>
g.
As
(
3
0
2
s)
+
6K1(aq)
6HCI(aq)
h.
2Na
(
3
0
S
2
aq) -F 2
(1
aq)
3F
(
2
g)
+
112(g)
8Cr
(
3
S
2
s)
—*
3NH
F
4
(s)
—
2Ag(s)
--
+
+ NF
(g)
3
H
S
2
(g)
203(g)
+
S
(
8
s)
-
8Na
(
3
0
S
2
aq)
76.
+
+
302(g)
4
S
2
21-1
(
aq)
0
+
-,
-+
—
PbCrO
(
4
s)
+
+
2NaCI(aq)
C0
(
2
g) + Na
(aq)
3
0
S
2
CuS0(aq) + 2
S0
(
g)
MnCI
(
2
aq)
—
2KN0
(
3
aq)
—
+
(‘12(g)
+
2H
O
2
(l)
±
2H
0
2
(l)
2As1
(
3
s) + 6KC1(aq) + 31120(l)
Na
(
6
0
4
S
2
aq)
+
2Nal(aq)
-
Copyright © Houghton Mifflin Company, All rights reserved.
L
-I--
-
93