Chapter 6, Problem 6.81QP

Interpretation Introduction

Interpretation: The amount of ammonium dichromate and potassium chlorate needed to make $\text{200}\text{.0}\text{L}$ of air at $\text{0}\text{.85}\text{atm}$ and $\text{273}\text{K}$ is to be calculated.

Concept introduction: The thermal decomposition of potassium chlorate generates oxygen and decomposition of ammonium dichromate generates pure nitrogen.

To determine: The amount of ammonium dichromate and potassium chlorate needed to make $\text{200}\text{.0}\text{L}$ of air at $\text{0}\text{.85}\text{atm}$ and $\text{273}\text{K}$.

Answer to Problem 6.81QP

Solution

The amount of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)$ required is $\underset{\_}{1.487\times 10^{3}g}$ and the amount of ${\text{KClO}}_3\left(s\right)$ required is $\underset{\_}{1.2\times 10^{2}g}$.

Explanation of Solution

Explanation

Given

The pressure is $\text{0}\text{.85}\text{atm}$.

The temperature is $\text{273}\text{K}$.

The volume of air is $\text{200}\text{.0}\text{L}$.

Air is about $78\%$ nitrogen by volume. Therefore, the volume of nitrogen is calculated as,

$\text{200}\text{.0}\text{L}\times \frac{78}{100}=\text{156}\text{L}$

Therefore, the volume of ${\text{N}}_{\text{2}}$ is $\text{156}\text{L}$.

Air is about $21\%$ oxygen by volume. Therefore, the volume of oxygen is calculated as,

$\text{200}\text{.0}\text{L}\times \frac{21}{100}=42\text{L}$

Therefore, the volume of ${\text{O}}_{\text{2}}$ is $\text{42}\text{L}$.

The amount of ${\text{N}}_{\text{2}}$ ( $n$) is calculated as,

$n=\frac{PV}{RT}$

Where,

$P$ is the pressure.

$V$ is the volume of ${\text{N}}_{\text{2}}$.

$R$ is gas constant $(0.08206l.atm/mol.K)$

$T$ is temperature.

Substitute the value of $P$, $V$, $R$ and $T$ in the above equation.

$\begin{array}{c}n=\frac{\text{0}\text{.85}\text{atm}\times \text{156}\text{L}}{0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{273}\text{K}}\\ =5.\text{9}\text{mol}\end{array}$

The amount of ${\text{O}}_{\text{2}}$ ( $n_1$) is calculated as,

$n_1=\frac{PV}{RT}$

Where,

$P$ is the pressure.

$V$ is the volume of ${\text{N}}_{\text{2}}$.

$R$ is gas constant $(0.08206l.atm/mol.K)$

$T$ is temperature.

Substitute the value of $P$, $V$, $R$ and $T$ in the above equation.

$\begin{array}{c}{n}_1=\frac{\text{0}\text{.85}\text{atm}\times 42\text{L}}{0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{273}\text{K}}\\ =1.5\text{mol}\end{array}$

The decomposition reaction of ammonium chromate is,

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(s\right)\to {\text{N}}_{\text{2}}\left(g\right)+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(g\right)$

According to the above reaction,

$\text{1}\text{mol}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)\approx \text{1}\text{mol}{\text{N}}_{\text{2}}\left(g\right)$

The molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)$ is $\text{252}\text{.07}\text{g/mol}$. Therefore, mass of $\text{1}\text{mol}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)$ is equal to $\text{252}\text{.07}\text{g}$.

Since $\text{1}\text{mol}{\text{N}}_{\text{2}}$ is produced from $\text{252}\text{.07}\text{g}$ of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)$.

Therefore, $5.\text{9}\text{mol}{\text{N}}_{\text{2}}$ is produced from $\begin{array}{c}\text{252}\text{.07}\times 5.\text{9}\text{g}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)\\ =\underset{\_}{1.487\times 10^{3}g}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)\end{array}$

Therefore, the amount of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)$ required is $\underset{\_}{1.487\times 10^{3}g}$.

The oxygen is generated by the thermal decomposition of potassium chlorate.

${\text{2KClO}}_3\left(s\right)\to \text{2KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

According to the above reaction,

$\text{2}\text{mol}{\text{KClO}}_3\left(s\right)\approx 3\text{mol}{\text{O}}_2\left(g\right)$

The molar mass of ${\text{KClO}}_3\left(s\right)$ is $\text{122}\text{.55}\text{g/mol}$. Therefore, mass of

$\begin{array}{c}2\text{mol}{\text{KClO}}_3\left(s\right)=2\times \text{122}\text{.55}\\ =\text{245}\text{.1}\text{g}\end{array}$

Since $3\text{mol}{\text{O}}_2\left(g\right)$ is produced from $\text{245}\text{.1}\text{g}$ of ${\text{KClO}}_3\left(s\right)$.

Therefore, $1.5\text{mol}{\text{O}}_{\text{2}}$ is produced from $\begin{array}{c}\frac{\text{245}\text{.1}\times 1.5}{3}\text{g}{\text{KClO}}_3\left(s\right)\\ =\underset{\_}{1.2\times 10^{2}g}{\text{KClO}}_3\left(s\right)\end{array}$

Therefore, the amount of ${\text{KClO}}_3\left(s\right)$ required is $\underset{\_}{1.2\times 10^{2}g}$.

Conclusion

The amount of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{s}\right)$ required is $\underset{\_}{1.487\times 10^{3}g}$ and the amount of ${\text{KClO}}_3\left(s\right)$ required is $\underset{\_}{1.2\times 10^{2}g}$