Chapter 4, Problem 4.70QP

(a)

Interpretation Introduction

Interpretation:

The given equation ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{ + CO}}_{\text{(g)}}\stackrel{}{\to }{\text{ Fe}}_{\text{3}}{\text{O}}_{\text{4(s)}}{\text{ + CO}}_{\text{2(g)}}$ has to be balanced.

Concept Introduction:

Balancing the equation:

• There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• First write the skeletal reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

Explanation of Solution

Given equation:

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{ + CO}}_{\text{(g)}}\stackrel{}{\to }{\text{ Fe}}_{\text{3}}{\text{O}}_{\text{4(s)}}{\text{ + CO}}_{\text{2(g)}}$

Balancing the chemical Equation:

Count the number of atoms on each side of the reaction.

Atom	Reactant side	Product side
$\text{Fe}$	2	3
$\text{O}$	4	6
$\text{C}$	1	1

Place suitable coefficient before reactants and products and then check for the number of atoms again.

  ${\text{3Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{ + CO}}_{\text{(g)}}\stackrel{}{\to }{\text{ 2Fe}}_{\text{3}}{\text{O}}_{\text{4(s)}}{\text{ + CO}}_{\text{2(g)}}$

Atom	Reactant side	Product side
$\text{Fe}$	6	6
$\text{O}$	10	10
$\text{C}$	1	1

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{3Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{ + CO}}_{\text{(g)}}\stackrel{}{\to }{\text{ 2Fe}}_{\text{3}}{\text{O}}_{\text{4(s)}}{\text{ + CO}}_{\text{2(g)}}$.

(b)

Interpretation Introduction

Interpretation:

The given equation ${\text{C}}_{\text{6}}{\text{H}}_{\text{6(l)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$ has to be balanced.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given equation:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{6(l)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Balancing the chemical Equation:

Count the number of atoms on each side of the reaction.

Atom	Reactant side	Product side
$\text{H}$	6	2
$\text{O}$	2	3
$\text{C}$	6	1

Place suitable coefficient before reactants and products and then check for the number of atoms again.

  ${\text{2C}}_{\text{6}}{\text{H}}_{\text{6(l)}}{\text{ + 15O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 12CO}}_{\text{2(g)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Atom	Reactant side	Product side
$\text{H}$	12	12
$\text{O}$	30	30
$\text{C}$	12	12

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{2C}}_{\text{6}}{\text{H}}_{\text{6(l)}}{\text{ + 15O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 12CO}}_{\text{2(g)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}$.

(c)

Interpretation Introduction

Interpretation:

The given equation ${\text{I}}_{\text{4}}{\text{O}}_{\text{9(s)}}\stackrel{}{\to }{\text{ I}}_{\text{2}}{\text{O}}_{\text{6(s)}}{\text{ + I}}_{\text{2(s)}}{\text{ + O}}_{\text{2(g)}}$ has to be balanced.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given equation:

  ${\text{I}}_{\text{4}}{\text{O}}_{\text{9(s)}}\stackrel{}{\to }{\text{ I}}_{\text{2}}{\text{O}}_{\text{6(s)}}{\text{ + I}}_{\text{2(s)}}{\text{ + O}}_{\text{2(g)}}$

Balancing the chemical Equation:

Count the number of atoms on each side of the reaction.

Atom	Reactant side	Product side
$\text{I}$	4	4
$\text{O}$	9	8

Place suitable coefficient before reactants and products and then check for the number of atoms again.

  ${\text{2I}}_{\text{4}}{\text{O}}_{\text{9(s)}}\stackrel{}{\to }{\text{ I}}_{\text{2}}{\text{O}}_{\text{6(s)}}{\text{ + 3I}}_{\text{2(s)}}{\text{ + 6O}}_{\text{2(g)}}$

Atom	Reactant side	Product side
$\text{I}$	8	8
$\text{O}$	18	18

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{2I}}_{\text{4}}{\text{O}}_{\text{9(s)}}\stackrel{}{\to }{\text{ I}}_{\text{2}}{\text{O}}_{\text{6(s)}}{\text{ + 3I}}_{\text{2(s)}}{\text{ + 6O}}_{\text{2(g)}}$.

(d)

Interpretation Introduction

Interpretation:

The given equation ${\text{KClO}}_{\text{3(s)}}\stackrel{}{\to }{\text{KCl}}_{\text{(s)}}{\text{ + O}}_{\text{2(g)}}$ has to be balanced.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given reaction:

  ${\text{KClO}}_{\text{3(s)}}\stackrel{}{\to }{\text{KCl}}_{\text{(s)}}{\text{ + O}}_{\text{2(g)}}$

Balancing the chemical Equation:

Count the number of atoms on each side of the reaction.

Atom	Reactant side	Product side
$\text{K}$	1	1
$\text{O}$	3	2
$\text{Cl}$	1	1

Place suitable coefficient before reactants and products and then check for the number of atoms again.

  ${\text{2KClO}}_{\text{3(s)}}\stackrel{}{\to }{\text{2KCl}}_{\text{(s)}}{\text{ + 3O}}_{\text{2(g)}}$

Atom	Reactant side	Product side
$\text{K}$	2	2
$\text{O}$	6	6
$\text{Cl}$	2	2

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{2KClO}}_{\text{3(s)}}\stackrel{}{\to }{\text{2KCl}}_{\text{(s)}}{\text{ + 3O}}_{\text{2(g)}}$.