Chapter 6, Problem 6.84QP

Interpretation Introduction

Interpretation: The amount of potassium nitrate and potassium chlorate needed to make $\text{200}\text{.0}\text{L}$ of gas at $1.00\text{atm}$ and $\text{290}\text{K}$ is to be calculated.

Concept introduction: The thermal decomposition of potassium chlorate generates oxygen. Nitrogen is produced from the reaction of sodium metal and potassium nitrate.

To determine: The amount of potassium nitrate and potassium chlorate needed to make $\text{200}\text{.0}\text{L}$ of gas at $1.00\text{atm}$ and $\text{290}\text{K}$.

Answer to Problem 6.84QP

Solution

The amount of ${\text{KNO}}_{\text{3}}\left(s\right)$ required is $\underset{\_}{1.358\times 10^{3}g}$ and the amount of ${\text{KClO}}_3\left(s\right)$ required is $\underset{\_}{1.37\times 10^{2}g}$.

Explanation of Solution

Explanation

Given

The pressure is $1.00\text{atm}$.

The temperature is $\text{290}\text{K}$.

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

The volume of ${\text{N}}_{\text{2}}$ is $\text{160}\text{L}$.

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.

$\begin{array}{c}n=\frac{1.00\text{atm}\times \text{160}\text{L}}{0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{290}\text{K}}\\ =6.72\text{mol}\end{array}$

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.

$\begin{array}{c}n=\frac{1.00\text{atm}\times 4\text{0}\text{L}}{0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{290}\text{K}}\\ =1.68\text{mol}\end{array}$

The chemical equation for the reaction of sodium metal and potassium nitrate to generate nitrogen is,

$\text{10Na}\left(s\right)+{\text{2KNO}}_{\text{3}}\left(s\right)\to {\text{K}}_{\text{2}}\text{O}\left(s\right)+{\text{5Na}}_{\text{2}}\text{O}\left(s\right)+{\text{N}}_{\text{2}}\left(g\right)$

According to the above reaction,

$2\text{mol}{\text{KNO}}_{\text{3}}\left(s\right)\approx \text{1}\text{mol}{\text{N}}_{\text{2}}\left(g\right)$

The molar mass of ${\text{KNO}}_{\text{3}}\left(s\right)$ is $\text{101}\text{.1}\text{g/mol}$. Therefore, mass of $\text{1}\text{mol}{\text{KNO}}_{\text{3}}\left(s\right)$ is equal to $\text{101}\text{.1}\text{g}$. Hence mass of $2\text{mol}{\text{KNO}}_{\text{3}}\left(s\right)$ is equal to $\begin{array}{c}\text{2}\times \text{101}\text{.1}\text{g}\\ =\text{202}\text{.2}\text{g}\end{array}$.

Since $\text{1}\text{mol}{\text{N}}_{\text{2}}$ is produced from $\text{202}\text{.2}\text{g}$ of ${\text{KNO}}_{\text{3}}\left(s\right)$.

Therefore, $6.72\text{mol}{\text{N}}_{\text{2}}$ is produced from $\begin{array}{c}6.72\times 202.2\text{g}{\text{KNO}}_{\text{3}}\left(s\right)\\ =\underset{\_}{1.358\times 10^{3}g}{\text{KNO}}_{\text{3}}\left(s\right)\end{array}$

Therefore, the amount of ${\text{KNO}}_{\text{3}}\left(s\right)$ required is $\underset{\_}{1.358\times 10^{3}g}$.

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, $1.68\text{mol}{\text{O}}_{\text{2}}$ is produced from $\begin{array}{c}\frac{\text{245}\text{.1}\times 1.68}{3}\text{g}{\text{KClO}}_3\left(s\right)\\ =\underset{\_}{1.37\times 10^{2}g}{\text{KClO}}_3\left(s\right)\end{array}$

Therefore, the amount of ${\text{KClO}}_3\left(s\right)$ required is $\underset{\_}{1.37\times 10^{2}g}$.

Conclusion

The amount of ${\text{KNO}}_{\text{3}}\left(s\right)$ required is $\underset{\_}{1.358\times 10^{3}g}$ and the amount of ${\text{KClO}}_3\left(s\right)$ required is $\underset{\_}{1.37\times 10^{2}g}$