Chapter 7, Problem 38A

(a)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $K{O}_2\left(s\right)+H_{2}O\left(l\right)\to KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

  $2K{O}_2\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

Explanation of Solution

  $K{O}_2\left(s\right)+H_{2}O\left(l\right)\to KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

First, list the number of atoms of each element involved in the reaction.

Element	Reactant side	Product side
K	1	1
O	3	5
H	2	3

Reactants side:

K = 1

O = 2 + 1 = 3

H = 2

Products side:

K = 1

O = 1 + 2 + 2 = 5

H = 1 + 2 = 3

Balance the number of H atoms on both sides. On the reactant side there are 2 H atoms and on the product side there are 3 H atoms. Put a 2 before H₂O and a 2 before KOH.

  $K{O}_2\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

Element	Reactant side	Product side
K	1	2
O	4	6
H	4	4

Reactants side:

K = 1

O = 2 + 1(2) = 4

H = 2(2) = 4

Products side:

K = 1(2) = 2

O = 1(2) + 2 + 2 = 6

H = 1(2) + 2 = 4

Balance the number of O atoms on both sides. On the reactant side there are 4 O atoms and on the product side there are 6 O atoms. On the reactant side put a 2 before KO₂

  $2K{O}_2\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

Element	Reactant side	Product side
K	2	2
O	6	6
H	4	4

Reactants side:

K = 1(2) = 2

O = 2(4) + 1(2) = 6

H = 2(2) = 4

Products side:

K = 1(2) = 2

O = 1(2) + 2 + 2 = 6

H = 1(2) + 2 = 4

The balanced equation is as follows:

  $2K{O}_2\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+O_2\left(g\right)+H_2{O}_2\left(aq\right)$

(b)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $F{e}_2{O}_3\left(s\right)+HN{O}_3\left(aq\right)\to Fe{(N{O}_3)}_3\left(aq\right)+H_{2}O\left(l\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

. $F{e}_2{O}_3\left(s\right)+6HN{O}_3\left(aq\right)\to 2 Fe{\left(N{O}_3\right)}_3\left(aq\right)+ 3 H_{2}O(l)$

Explanation of Solution

  $F{e}_2{O}_3\left(s\right)+HN{O}_3\left(aq\right)\to Fe{(N{O}_3)}_3\left(aq\right)+H_{2}O\left(l\right)$

First, list the number of atoms of each element involved in the reaction.

Element	Reactant side	Product side
P	2	1
Cl	6	10
H	1	2
O	1	3

Reactants side:

Fe = 2

O = 3 + 3 = 6

H = 1

N = 1

Products side:

Fe = 1

O = 9 + 1 = 10

H = 2

N = 3

Balance the number of Fe atoms on both sides. On the reactant side there are 2 Fe atoms and on the product side there are 1 Fe atom. Put a 2 before Fe(NO₃)₃

  $\begin{array}{l}F{e}_2{O}_3\left(s\right)+HN{O}_3\left(aq\right)\to 2Fe{\left(N{O}_3\right)}_3\left(aq\right)+H_{2}O\left(l\right)\hfill \\ \hfill \end{array}$

Element	Reactant side	Product side
P	2	2
Cl	6	19
H	1	2
O	1	6

Reactants side:

Fe = 2

O = 3 + 3 = 6

H = 1

N = 1

Products side:

Fe = 1(2) = 2

O = 9(2) + 1 = 19

H = 2

N = 3(2) = 6

Balance the number of N atoms on both sides. On the reactant side there are 1 N atoms and on the product side there are 6 N atoms. Put a 6 before HNO₃

  $F{e}_2{O}_3\left(s\right)+6HN{O}_3\left(aq\right)\to 2Fe{\left(N{O}_3\right)}_3\left(aq\right)+H_{2}O\left(l\right)$

Element	Reactant side	Product side
P	2	2
Cl	21	19
H	6	2
O	6	6

Reactants side:

Fe = 2

O = 3 + 3(6) = 21

H = 1(6) = 6

N = 1(6) = 6

Products side:

Fe = 1(2) = 2

O = 9(2) + 1 = 19

H = 2

N = 3(2) = 6

Balance the number of H atoms on both sides. On the reactant side there are 6 H atoms and on the product side there are 2 H atoms. Put a 3 before H₂O

  $F{e}_2{O}_3\left(s\right)+6HN{O}_3\left(aq\right)\to 2Fe{\left(N{O}_3\right)}_3\left(aq\right)+3H_{2}O\left(l\right)$

Element	Reactant side	Product side
Fe	2	2
O	21	21
H	6	6
N	6	6

Reactants side:

Fe = 2

O = 3 + 3(6) = 21

H = 1(6) = 6

N = 1(6) = 6

Products side:

Fe = 1(2) = 2

O = 9(2) + 1(3) = 21

H = 2(3) = 6

N = 3(2) = 6

The equation is already balanced:

  $F{e}_2{O}_3\left(s\right)+6HN{O}_3\left(aq\right)\to 2 Fe{\left(N{O}_3\right)}_3\left(aq\right)+ 3 H_{2}O(l)$

(c)

Interpretation Introduction

Interpretation:

The given chemical equation is to be balanced.

  $N{H}_3\left(g\right)+O_2\left(g\right)\to NO\left(g\right)+H_{2}O\left(g\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

  $4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$

Explanation of Solution

  $N{H}_3\left(g\right)+O_2\left(g\right)\to NO\left(g\right)+H_{2}O\left(g\right)$

First, list the number of atoms of each element involved in the reaction.

Element	Reactant side	Product side
N	1	1
H	3	2
O	2	2

Reactants side:

N = 1

H = 3

O = 2

Products side:

N = 1

H = 2

O = 1 + 1 = 2

Balance the number of H atoms on both sides. On the reactant side there are 3 H atoms and on the product side there are 2 H atom. Put a 2 before NH₃ and put a 3 before H₂O

  $\begin{array}{l}2N{H}_3\left(g\right)+O_2\left(g\right)\to NO\left(g\right)+3H_{2}O\left(g\right)\hfill \\ \hfill \end{array}$

Element	Reactant side	Product side
N	2	1
H	6	6
O	2	4

Reactants side:

N = 1(2) = 2

H = 3(2) = 6

O = 2

Products side:

N = 1

H = 2(3) = 6

O = 1 + 1(3) = 4

Balance the number of N atoms on both sides. On the reactant side there are 2 N atoms and on the product side there are 1 N atoms. Put a 2 before NO.

  $2N{H}_3\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)+3H_{2}O\left(g\right)$

Element	Reactant side	Product side
N	2	2
H	6	6
O	2	5

Reactants side:

N = 1(2) = 2

H = 3(2) = 6

O = 2

Products side:

N = 1(2) = 2

H = 2(3) = 6

O = 1(2) + 1(3) = 5

Balance the number of O atoms on both sides. On the reactant side there are 2 O atoms and on the product side there are 5 atoms. Put a 5/2 before O₂

  $2N{H}_3\left(g\right)+5/2O_2\left(g\right)\to 2NO\left(g\right)+3H_{2}O\left(g\right)$

Element	Reactant side	Product side
N	2	2
H	6	6
O	5	5

Reactants side:

N = 1(2) = 2

H = 3(2) = 6

O = 2(5/2) = 5

Products side:

N = 1(2) = 2

H = 2(3) = 6

O = 1(2) + 1(3) = 5

Multiply the equation by 2 to get the lowest ratio of whole numbers.

  $2\times \left[2N{H}_3\left(g\right)+5/2O_2\left(g\right)\to 2NO\left(g\right)+3H_{2}O\left(g\right)\right]$

The equation is already balanced:

  $4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$

(d)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $PC{l}_5\left(l\right)+H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+HCl\left(g\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

  $PC{l}_5\left(l\right)+4H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+5HCl\left(g\right)$

Explanation of Solution

  $PC{l}_5\left(l\right)+H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+HCl\left(g\right)$

First, list the number of atoms of each element involved in the reaction.

Element	Reactant side	Product side
P	1	1
Cl	5	1
H	2	4
O	1	4

Reactants side:

P = 1

Cl = 5

H = 2

O = 1

Products side:

P = 1

Cl = 1

H = 3 + 1

O = 4

Balance the number of Cl atoms on both sides. On the reactant side there are 5 Cl atoms and on the product side there are 1 Cl atom. Put a 5 before HCl.

  $\begin{array}{l}PC{l}_5\left(l\right)+H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+5HCl\left(g\right)\hfill \\ \hfill \end{array}$

Element	Reactant side	Product side
P	1	1
Cl	5	5
H	2	8
O	1	4

Reactants side:

P = 1

Cl = 5

H = 2

O = 1

Products side:

P = 1

Cl = 1(5) = 5

H = 3 + 1(5) = 8

O = 4

Balance the number of H atoms on both sides. On the reactant side there are 2 H atoms and on the product side there are 8 H atoms. Put a 4 before H₂O.

  $PC{l}_5\left(l\right)+4H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+5HCl\left(g\right)$

Element	Reactant side	Product side
P	1	1
Cl	5	5
H	8	8
O	4	4

Reactants side:

P = 1

Cl = 5

H = 2(4) = 8

O = 1(4) = 4

Products side:

P = 1

Cl = 1(5) = 5

H = 3 + 1(5) = 8

O = 4

The equation is already balanced:

  $PC{l}_5\left(l\right)+4H_{2}O\left(l\right)\to H_{3}P{O}_4\left(aq\right)+5HCl\left(g\right)$

(e)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $C_2{H}_{5}OH\left(l\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

  $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

Explanation of Solution

  $C_2{H}_{5}OH\left(l\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$

First, list the number of atoms of each element involved in the reaction.

Element	Reactant side	Product side
C	2	1
H	6	2
O	3	3

Balance the number of C atoms on both sides. On the reactant side there are 2 C atoms and on the product side there are 1 C atom. Put a 2 before CO₂

  $C_2{H}_{5}OH\left(l\right)+O_2\left(g\right)\to 2C{O}_2\left(g\right)+H_{2}O\left(l\right)$

Element	Reactant side	Product side
C	2	2
H	6	2
O	3	5

Balance the number of H atoms on both sides. On the reactant side there are 6 H atoms and on the product side there are 2 H atoms. Put a 3 before H₂O.

  $C_2{H}_{5}OH\left(l\right)+O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

Element	Reactant side	Product side
C	2	2
H	6	6
O	3	7

Balance the number of O atoms on both sides. On the reactant side there are 3 O atoms and on the product side there are 7 O atoms. Put a 3 before O₂.

  $\begin{array}{l}{C}_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)\hfill \\ \hfill \end{array}$

Element	Reactant side	Product side
C	2	2
H	6	6
O	7	7

The equation is already balanced:

  $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

(f)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $CaO\left(s\right)+C\left(s\right)\to Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

  $\begin{array}{l}2CaO\left(s\right)+5C\left(s\right)\to 2Ca{C}_2\left(s\right)+C{O}_2\left(g\right)\hfill \\ \hfill \end{array}$

Explanation of Solution

First, list the number of atoms of each element involved in the reaction.

  $CaO\left(s\right)+C\left(s\right)\to Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

Element	Reactant side	Product side
Ca	1	1
O	1	2
C	1	3

Balance the number of O atoms on both sides. On the reactant side there are 1 O atom and on the product side there are 2 O atoms. Put a 2 before CaO(s).

  $2CaO\left(s\right)+C\left(s\right)\to Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

Element	Reactant side	Product side
Ca	2	1
O	2	2
C	1	3

Balance the number of Ca atoms on both sides. On the reactant side there are 2 Ca atoms and on the product side there are 1 Ca atom. Put a 2 before CaC₂.

  $2CaO\left(s\right)+C\left(s\right)\to 2Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

Element	Reactant side	Product side
Ca	2	2
O	2	2
C	1	5

Balance the number of C atoms on both sides. On the reactant side there are 1 C atom and on the product side there are 5 C atoms. Put a 5 before C(s).

  $2CaO\left(s\right)+5C\left(s\right)\to 2Ca{C}_2\left(s\right)+C{O}_2\left(g\right)$

Element	Reactant side	Product side
Ca	2	2
O	2	2
C	5	5

The equation is already balanced

  $\begin{array}{l}2CaO\left(s\right)+5C\left(s\right)\to 2Ca{C}_2\left(s\right)+C{O}_2\left(g\right)\hfill \\ \hfill \end{array}$

(g)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $Mo{S}_2\left(s\right)+O_2\left(g\right)\to Mo{O}_3\left(s\right)+S{O}_2\left(g\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

Explanation of Solution

First, list the number of atoms of each element involved in the reaction.

  $Mo{S}_2\left(s\right)+O_2\left(g\right)\to Mo{O}_3\left(s\right)+S{O}_2\left(g\right)$

Reactants side:

Mo = 1

S = 2

O = 2

Products side:

Mo = 1

S = 1

O = 3 + 2 = 5

Balance the number of S atoms on both sides. On the reactant side there are 2 S atoms and on the product side there are 1 S atom. Put a 2 before SO₂.

  $Mo{S}_2\left(s\right)+O_2\left(g\right)\to Mo{O}_3\left(s\right)+2S{O}_2\left(g\right)$

Element	Reactant side	Product side
Mo	1	1
S	2	2
O	2	7

Balance the number of O atoms on both sides. On the reactant side there are 2 O atoms and on the product side there are 7 O atoms. Put a 7/2 before O₂.

  $Mo{S}_2\left(s\right)+7/2O_2\left(g\right)\to Mo{O}_3\left(s\right)+2S{O}_2\left(g\right)$

Multiply the equation by 2 to get the lowest ratio of whole numbers.

  $\begin{array}{l}2\times [Mo{S}_2\left(s\right)+7/2O_2\left(g\right)\to Mo{O}_3\left(s\right)+2S{O}_2\left(g\right)]\hfill \\ \hfill \end{array}$

Element	Reactant side	Product side
Mo	1	1
S	2	2
O	7	7

The equation is already balanced:

  $2Mo{S}_2\left(s\right)+7O_2\left(g\right)\to 2Mo{O}_3\left(s\right)+4S{O}_2\left(g\right)$

(h)

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced.

  $FeC{O}_3\left(s\right)+H_{2}C{O}_3\left(aq\right)\to Fe{(HC{O}_3)}_2\left(aq\right)$

Concept Introduction:

First, list the number of atoms of each element involved in the reaction then balance the given chemical equation.

Answer to Problem 38A

  $FeC{O}_3\left(s\right)+H_{2}C{O}_3\left(aq\right)\to Fe{(HC{O}_3)}_2\left(aq\right)$

Explanation of Solution

First, list the number of atoms of each element involved in the reaction.

  $FeC{O}_3\left(s\right)+H_{2}C{O}_3\left(aq\right)\to Fe{(HC{O}_3)}_2\left(aq\right)$

Element	Reactant side	Product side
Fe	1	1
C	2	2
H	2	2
O	6	6

The equation is already balanced:

  $FeC{O}_3\left(s\right)+H_{2}C{O}_3\left(aq\right)\to Fe{(HC{O}_3)}_2\left(aq\right)$