Chapter 6, Problem 18QP

(a)

Interpretation Introduction

Interpretation:

The mole ratio of each product to the reactant ${\text{NH}}_4{\text{NO}}_3$ is to be determined.

Explanation of Solution

The balanced equation of the given reaction is:

$2{\text{NH}}_4{\text{NO}}_3\left(s\right)\to 2{\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(g\right)$

It is apparent from the equation that $2\text{ mole}$ of nitrogen, $1\text{ mole}$ of oxygen, and $4\text{ mole}$ of water are formed when $2\text{ mole}$ of ammonium nitrate react. Therefore, the required mole ratios for nitrogen and oxygen to ammonium nitrate are:

$\begin{array}{c}\frac{\text{moles N}}{{\text{moles NH}}_4{\text{NO}}_3}=\frac{2}{2}\\ =1\\ \frac{{\text{moles O}}_2}{{\text{moles NH}}_4{\text{NO}}_3}=\frac{1}{2}\\ =0.5\end{array}$

The required mole ratios for water to ammonium nitrate are:

$\begin{array}{c}\frac{{\text{moles H}}_2\text{O}}{{\text{moles NH}}_4{\text{NO}}_3}=\frac{4}{2}\\ =2\end{array}$

(b)

Interpretation Introduction

Interpretation:

The number of moles of each product formed when $1$ mole of ammonium nitrate reacts is to be determined.

Explanation of Solution

The balanced equation of the given reaction is:

$2{\text{NH}}_4{\text{NO}}_3\left(s\right)\to 2{\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(g\right)$

It is apparent from the equation that $2\text{ mole}$ of nitrogen, $1\text{ mole}$ of oxygen, and $4\text{ mole}$ of water are formed when $2\text{ mole}$ of ammonium nitrate react. If the reaction starts from $1$ mole of ammonium nitrate, then for nitrogen,

$\begin{array}{c}2{\text{ mole NH}}_4{\text{NO}}_3=2\text{ mole N}\\ 1{\text{mole NH}}_4{\text{NO}}_3=\frac{2}{2}\text{ mole N}\\ =1\text{ mole N}\end{array}$

Similarly, for oxygen,

$\begin{array}{c}2{\text{ mole NH}}_4{\text{NO}}_3=1{\text{ mole O}}_2\\ 1{\text{mole NH}}_4{\text{NO}}_3=\frac{1}{2}{\text{ mole O}}_2\\ =0.5{\text{ mole O}}_2\end{array}$

Similarly, for water,

$\begin{array}{c}2{\text{ mole NH}}_4{\text{NO}}_3=4{\text{ mole H}}_2\text{O}\\ 1{\text{mole NH}}_4{\text{NO}}_3=\frac{4}{2}{\text{ mole H}}_2\text{O}\\ =2{\text{ mole H}}_2\text{O}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of each product formed when $6.25$ moles of ammonium nitrate react is to be determined.

Explanation of Solution

The balanced equation of the given reaction is:

$2{\text{NH}}_4{\text{NO}}_3\left(s\right)\to 2{\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(g\right)$

It is apparent from the equation that $2\text{ mole}$ of nitrogen, $1\text{ mole}$ of oxygen, and $4\text{ mole}$ of water are formed when $2\text{ mole}$ of ammonium nitrate react. If the reaction starts from $12.4\text{ mole}$ of ammonium nitrate, then for nitrogen.

$\begin{array}{c}2{\text{ mole NH}}_4{\text{NO}}_3=2\text{ mole N}\\ 1{\text{mole NH}}_4{\text{NO}}_3=\frac{2}{2}\text{ mole N}\\ 12.4{\text{mole NH}}_4{\text{NO}}_3=\frac{2}{2}\times \text{12}\text{.4 mole N}\\ =12.4\text{ mole N}\end{array}$

Similarly, for oxygen,

$\begin{array}{c}2{\text{ mole NH}}_4{\text{NO}}_3=1{\text{ mole O}}_2\\ 1{\text{mole NH}}_4{\text{NO}}_3=\frac{1}{2}{\text{ mole O}}_2\\ 12.4{\text{mole NH}}_4{\text{NO}}_3=\frac{1}{2}\times \text{12}{\text{.4 mole O}}_2\\ =6.2{\text{ mole O}}_2\end{array}$

Similarly, for water,

$\begin{array}{c}2{\text{ mole NH}}_4{\text{NO}}_3=4{\text{ mole H}}_2\text{O}\\ 1{\text{mole NH}}_4{\text{NO}}_3=\frac{4}{2}{\text{ mole H}}_2\text{O}\\ 12.4{\text{mole NH}}_4{\text{NO}}_3=\frac{4}{2}\times \text{12}{\text{.4 mole H}}_2\text{O}\\ =24.8{\text{ mole H}}_2\text{O}\end{array}$