Chapter 4, Problem 1E

Balance the following equations by inspection.
a. ${\text{SO}}_{\text{2}}\to {\text{SO}}_{\text{2}}{\text{+ O}}_{\text{2}}$
b. ${\text{CI}}_{\text{2}}{\text{O}}_{\text{7}}{\text{ + H}}_{\text{2}}\text{O}\to {\text{HCIO}}_{\text{4}}$
c. ${\text{NO}}_{\text{2}}{\text{+ H}}_{\text{2}}\text{O}\to {\text{HNO}}_{\text{2}}\text{+NO}$
d. ${\text{PCI}}_{\text{2}}{\text{+ H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}\text{+ HCI}$

Interpretation Introduction

(a)

Interpretation:

Balance the following reaction by using inspection method.

${\text{SO}}_{\text{3}}\stackrel{}{\to }{\text{SO}}_{\text{2}}{\text{+ O}}_{\text{2}}$

Concept introduction:

Points for balancing the chemical equation:

• First, balance an element which occurs in only one compound present on both sides of the equation. Use fractions if required to balance an element.
• Then balance the atoms of element present in more amounts.
• In order to make fractions into whole number, use multiplication with suitable coefficient.

Answer to Problem 1E

The balanced equation is

${\text{2SO}}_{\text{3}}\stackrel{}{\to }{\text{2SO}}_{\text{2}}{\text{+ O}}_{\text{2}}$

Explanation of Solution

Given information:

The given equation is

${\text{SO}}_{\text{3}}\stackrel{}{\to }{\text{SO}}_{\text{2}}{\text{+ O}}_{\text{2}}$

Since, less number of atoms of sulfur is present in the equation so first balance sulfur atoms.

As per equation, number of sulfur atoms is equal on both sides.

Now, balance number of oxygen atoms.

To balance the oxygen atoms, multiply the oxygen molecule with half.

${\text{SO}}_{\text{3}}\stackrel{}{\to }{\text{SO}}_{\text{2}}\text{+ }\frac{1}{2}{\text{O}}_{\text{2}}$

Multiply the whole equation with “2” to get co-efficient in whole number.

2× [ ${\text{SO}}_{\text{3}}\stackrel{}{\to }{\text{SO}}_{\text{2}}\text{+ }\frac{1}{2}{\text{O}}_{\text{2}}$ ]

Thus, balanced equation is given by:

${\text{2SO}}_{\text{3}}\stackrel{}{\to }{\text{2SO}}_{\text{2}}{\text{+ O}}_{\text{2}}$

Interpretation Introduction

(b)

Interpretation:

Balance the following reaction by using inspection method.

${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}{\text{+ H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{HClO}}_{\text{4}}$

Concept introduction:

Points for balancing the chemical equation:

• First, balance an element which occurs in only one compound present on both sides of the equation. Use fractions if required to balance an element.
• Then balance the atoms of element present in more amounts.
• In order to make fractions into whole number, use multiplication with suitable coefficient.

Answer to Problem 1E

The balanced chemical equation is:

${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}{\text{+ H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2HClO}}_{\text{4}}$

Explanation of Solution

Given information:

The given equation is:

${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}{\text{+ H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{HClO}}_{\text{4}}$

The more complex compound is Cl₂O₇, which has less common chlorine atom in it. So, balance chlorine atoms first.

The reactant side has two chlorine atoms and product has one chlorine atoms, so multiply ${\text{HClO}}_{\text{4}}$ by 2. The equation is given by:

${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}{\text{+ H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2HClO}}_{\text{4}}$

The above equation is balanced as number of oxygen and hydrogen atoms are equal on both sides.

Interpretation Introduction

(c)

Interpretation:

Balance the following reaction by using inspection method.

${\text{NO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{HNO}}_{\text{3}}\text{+ NO}$

Concept introduction:

Points for balancing the chemical equation:

• First, balance an element which occurs in only one compound present on both sides of the equation. Use fractions if required to balance an element.
• Then balance the atoms of element present in more amounts.
• In order to make fractions into whole number, use multiplication with suitable coefficient.

Answer to Problem 1E

The balanced chemical equation is:

${\text{3NO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2HNO}}_{\text{3}}\text{+ NO}$

Explanation of Solution

Given information:

The given equation is

${\text{NO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{HNO}}_{\text{3}}\text{+ NO}$

Since, least common atom is hydrogen atom so, first balance it on both the sides.

To balance hydrogen atoms, multiply ${\text{HNO}}_{\text{3}}$ with two. The equation is given by:

${\text{NO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2HNO}}_{\text{3}}\text{+ NO}$

Now, balance the number of nitrogen atoms on both the sides by multiplying ${\text{NO}}_{\text{2}}$ with three.

${\text{3NO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2HNO}}_{\text{3}}\text{+ NO}$

Thus, above equation is balanced.

Interpretation Introduction

(d)

Interpretation:

Balance the following reaction by using inspection method.

${\text{PCl}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{+ HCl}$

Concept introduction:

Points for balancing the chemical equation:

• First, balance an element which occurs in only one compound present on both sides of the equation. Use fractions if required to balance an element.
• Then balance the atoms of element present in more amounts.
• In order to make fractions into whole number, use multiplication with suitable coefficient.

Answer to Problem 1E

The balanced equation is

${\text{PCl}}_{\text{3}}{\text{+ 3H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{+ 3HCl}$

Explanation of Solution

Given information:

The given equation is

${\text{PCl}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{+ HCl}$

The least common atom is chlorine so; first balance it on both sides.

To balance chlorine atoms, multiply $\text{HCl}$ with three. The equation is given by:

${\text{PCl}}_{\text{3}}{\text{+ H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{+ 3HCl}$

Now, balance hydrogen on both the sides by multiplying ${\text{H}}_{\text{2}}\text{O}$ with three. The equation is given by:

${\text{PCl}}_{\text{3}}{\text{+ 3H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{+ 3HCl}$

Thus, above equation is balanced.