Chapter 6, Problem 17QP

(a)

Interpretation Introduction

Interpretation:

The mole ratio of each product to the reactant ${\text{C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3$ is to be determined.

Explanation of Solution

The balanced equation of the given reaction is:

$2{\text{C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3\left(s\right)\to 7\text{C}\left(s\right)+7\text{CO}\left(g\right)+3{\text{N}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$

It is apparent from the equation that $7\text{ mol}$ of carbon and carbon monoxide, $3\text{ mole}$ of nitrogen, and $5\text{ mole}$ of water are formed when $2\text{ mole}$ of TNT react. Therefore, the required mole ratios for carbon and carbon monoxide to TNT are:

$\begin{array}{c}\frac{\text{moles C}}{{\text{moles C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3}=\frac{7}{2}\\ =3.5\\ \frac{\text{moles CO}}{{\text{moles C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3}=\frac{7}{2}\\ =3.5\end{array}$

The required mole ratios for nitrogen and water to TNT are:

$\begin{array}{c}\frac{\text{moles N}}{{\text{moles C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3}=\frac{3}{2}\\ =1.5\\ \frac{{\text{moles H}}_2\text{O}}{{\text{moles C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3}=\frac{5}{2}\\ =2.5\end{array}$

(b)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The balanced equation of the given reaction is:

$2{\text{C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3\left(s\right)\to 7\text{C}\left(s\right)+7\text{CO}\left(g\right)+3{\text{N}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$

It is apparent from the equation that $7\text{ mol}$ of carbon and carbon monoxide, $3\text{ mole}$ of nitrogen, and $5\text{ mole}$ of water are formed when $2\text{ mole}$ of TNT react. If the reaction starts from $1$ mole of TNT, then for carbon,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=7\text{ mole C}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{7}{2}\text{ mole C}\\ =3.5\text{ mole C}\end{array}$

Similarly, for carbon monoxide,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=7\text{ mole CO}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{7}{2}\text{ mole CO}\\ =3.5\text{ mole CO}\end{array}$

Similarly, for nitrogen,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=3\text{ mole N}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{3}{2}\text{ mole N}\\ =1.5\text{ mole N}\end{array}$

Similarly, for water,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=5{\text{ mole H}}_2\text{O}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{5}{2}{\text{ mole H}}_2\text{O}\\ =2.5{\text{ mole H}}_2\text{O}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of each product formed when $6.25\text{ mole}$ of TNT react is to be determined.

Explanation of Solution

The balanced equation of the given reaction is:

$2{\text{C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3\left(s\right)\to 7\text{C}\left(s\right)+7\text{CO}\left(g\right)+3{\text{N}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$

It is apparent from the equation that $7\text{ mol}$ of carbon and carbon monoxide, $3\text{ mole}$ of nitrogen, and $5\text{ mole}$ of water are formed when $2\text{ mole}$ of TNT react. If the reaction starts from $6.25\text{ mole}$ of TNT, then for carbon,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=7\text{ mole C}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{7}{2}\text{ mole C}\\ 6.25{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{7}{2}\times \text{6}\text{.25 mole C}\\ =21.87\text{ mole C}\end{array}$

Similarly, for carbon monoxide,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=7\text{ mole CO}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{7}{2}\text{ mole CO}\\ 6.25{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{7}{2}\times 6.25\text{ mole CO}\\ =21.87\text{ mole CO}\end{array}$

Similarly, for nitrogen,

a

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=3\text{ mole N}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{3}{2}\text{ mole N}\\ 6.25{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{3}{2}\times 6.25\text{ mole N}\\ =9.37\text{ mole N}\end{array}$

Similarly, for water,

$\begin{array}{c}2{\text{ mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=5{\text{ mole H}}_2\text{O}\\ 1{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{5}{2}{\text{ mole H}}_2\text{O}\\ 6.25{\text{mole C}}_7{\text{H}}_5{\left({\text{NO}}_2\right)}_3=\frac{5}{2}\times 6.25{\text{ mole H}}_2\text{O}\\ =15.62{\text{ mole H}}_2\text{O}\end{array}$