Chapter 4, Problem 4.65QP

(a)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

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:

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

Balancing the chemical Equation:

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

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

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

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6(g)}}{\text{ + 7O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 4CO}}_{\text{2(g)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Atom	Reactant side	Product side
$\text{C}$	4	4
$\text{H}$	12	12
$\text{O}$	14	14

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6(g)}}{\text{ + 7O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 4CO}}_{\text{2(g)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}$.

(b)

Interpretation Introduction

Interpretation:

The given equation ${\text{K}}_{\text{2}}{\text{O}}_{\text{(s)}}{\text{ + P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\stackrel{}{\to }\text{ K}{\text{3}}_{}{\text{PO}}_{\text{4(s)}}$ has to be balanced.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given equation:

  ${\text{K}}_{\text{2}}{\text{O}}_{\text{(s)}}{\text{ + P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\stackrel{}{\to }\text{ K}{\text{3}}_{}{\text{PO}}_{\text{4(s)}}$

Balancing the chemical Equation:

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

Atom	Reactant side	Product side
$\text{K}$	2	3
$\text{O}$	11	4
$\text{P}$	4	1

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

  ${\text{6K}}_{\text{2}}{\text{O}}_{\text{(s)}}{\text{ + P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\stackrel{}{\to }\text{ 4K}{\text{3}}_{}{\text{PO}}_{\text{4(s)}}$

Atom	Reactant side	Product side
$\text{K}$	12	12
$\text{O}$	16	16
$\text{P}$	4	4

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{6K}}_{\text{2}}{\text{O}}_{\text{(s)}}{\text{ + P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\stackrel{}{\to }\text{ 4K}{\text{3}}_{}{\text{PO}}_{\text{4(s)}}$.

(c)

Interpretation Introduction

Interpretation:

The given equation ${\text{MgBr}}_{\text{2(aq)}}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ HBr}}_{\text{(g)}}{\text{ + MgSO}}_{\text{4(aq)}}$ has to be balanced.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given equation:

  ${\text{MgBr}}_{\text{2(aq)}}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ HBr}}_{\text{(g)}}{\text{ + MgSO}}_{\text{4(aq)}}$

Balancing the chemical Equation:

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

Atom	Reactant side	Product side
$\text{Mg}$	1	1
$\text{Br}$	2	1
$\text{H}$	2	1
$\text{S}$	1	1
$\text{O}$	4	4

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

  ${\text{MgBr}}_{\text{2(aq)}}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ 2HBr}}_{\text{(g)}}{\text{ + MgSO}}_{\text{4(aq)}}$

Atom	Reactant side	Product side
$\text{Mg}$	1	1
$\text{Br}$	2	2
$\text{H}$	2	2
$\text{S}$	1	1
$\text{O}$	4	4

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{MgBr}}_{\text{2(aq)}}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ 2HBr}}_{\text{(g)}}{\text{ + MgSO}}_{\text{4(aq)}}$.

(d)

Interpretation Introduction

Interpretation:

The given equation ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$ has to be balanced.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given reaction:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2(g)}}\stackrel{}{\to }{\text{CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Balancing the chemical Equation:

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

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

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

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + 3O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2CO}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Atom	Reactant side	Product side
$\text{C}$	2	2
$\text{H}$	6	6
$\text{O}$	7	7

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + 3O}}_{\text{2(g)}}\stackrel{}{\to }{\text{2CO}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(g)}}$.