Chapter 10, Problem 60QP

Interpretation Introduction

Interpretation:

The mass of solid ${\text{NH}}_4\text{Cl}$ and the volume of gas left after the completion of reaction are to be calculated and that gas is to be identified.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and the number of moles with each other with the help of gas laws.

This can be shown by:

$P=\frac{nRT}{V}$

Here, R represents the universal gas constant, $n$ is the number of moles, $T$ is the temperature, $P$ is the pressure, and $V$ is the volume.

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Number of moles is defined as the ratio of the mass to the molar mass.

$n=\frac{m}{M}$

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

Answer to Problem 60QP

Solution: $107\text{ g}$ and $54.5\text{ L}$, the gas is ${\text{NH}}_3$.

Explanation of Solution

Given information:

Mass: $\begin{array}{l}{m}_{{\text{NH}}_3}=73\text{ g}\\ m_{\text{HCl}}=73\text{ g}\end{array}$

Temperature $T=14°\text{C}$

Pressure $P=752\text{ mmHg}$

The reaction for the formation of ${\text{NH}}_4\text{Cl}$ is as follows:

${\text{NH}}_3\left(g\right)+\text{HCl}\left(g\right)\to {\text{NH}}_4\text{Cl}\left(g\right)$

The molar mass of ${\text{NH}}_3$ is $17\text{ g}/\text{mol}$.

Calculate the number of moles of ${\text{NH}}_3$ as follows:

$n_{{\text{NH}}_{\text{3}}}=\frac{m_{{\text{NH}}_{\text{3}}}}{M_{{\text{NH}}_{\text{3}}}}$

Substitute $73\text{ g}$ for $m_{{\text{NH}}_{\text{3}}}$ and $17\text{ g}/\text{mol}$ for $M_{{\text{NH}}_{\text{3}}}$ in the above equation

$\begin{array}{c}{n}_{{\text{NH}}_{\text{3}}}=\frac{73\text{ g}}{17\text{ g}/\text{mol}}\\ =4.29\text{ mol}\end{array}$

The molar mass of $\text{HCl}$ is $36\text{ g}/\text{mol}$.

Calculate the number of moles of $\text{HCl}$ as follows:

$n_{\text{HCl}}=\frac{m_{\text{HCl}}}{M_{\text{HCl}}}$

Substitute $73\text{ g}$ for $m_{\text{HCl}}$ and $36\text{ g}/\text{mol}$ for $M_{\text{HCl}}$ in the above equation

$\begin{array}{c}{n}_{\text{HCl}}=\frac{73\text{ g}}{36\text{ g}/\text{mol}}\\ =2\text{ mol}\end{array}$

Hydrochloric acid ($\text{HCl}$) is a limiting reactant because both ${\text{NH}}_3$ and $\text{HCl}$ react in the ratio of $1:1$.

The number of moles of ${\text{NH}}_4\text{Cl}$ formed is

$n_{{\text{NH}}_4\text{Cl}}=n_{\text{HCl}}$

Substitute $2\text{ mol}$ for $n_{\text{HCl}}$ in the above equation

$n_{{\text{NH}}_4\text{Cl}}=2\text{ mol}$

The molar mass of ${\text{NH}}_4\text{Cl}$ is $53.4\text{ g}/\text{mol}$.

Calculate the mass of ${\text{NH}}_4\text{Cl}$ as follows:

$m_{{\text{NH}}_4\text{Cl}}=n_{{\text{NH}}_4\text{Cl}}\times M_{{\text{NH}}_4\text{Cl}}$

Substitute $2\text{ mol}$ for $n_{{\text{NH}}_4\text{Cl}}$ and $53.4\text{ g}/\text{mol}$ for $M_{{\text{NH}}_4\text{Cl}}$ in the above equation

$\begin{array}{c}{m}_{{\text{NH}}_4\text{Cl}}=2\text{ mol}\times 53.4\text{ g}/\text{mol}\\ =106.8\text{ g}\\ \approx 107\text{ g}\end{array}$

So, the gas left after the completion of the reaction is ${\text{NH}}_3$ and its moles are

$n_{{\text{NH}}_3}\left(\text{remaining}\right)=n_{{\text{NH}}_3}-n_{\text{HCl}}$

Substitute $4.29\text{ mol}$ for $n_{{\text{NH}}_3}$ and $\text{2 mol}$ for $n_{\text{HCl}}$ in the above equation

$\begin{array}{c}{n}_{{\text{NH}}_3}\left(\text{remaining}\right)=4.29\text{ mol}-2\text{ mol}\\ =2.29\text{ mol}\end{array}$

The temperature is14°C.

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula as

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(14+273.15\right)\\ =287.15\text{ K}\end{array}$

The pressure is $752\text{ mmHg}$.

Convert mmHg to atm:

$\begin{array}{c}752\text{ mmHg}=\left(752\cancel{\text{mmHg}}\right)\left(\frac{1/760\text{ atm}}{\text{1 }\cancel{\text{mmHg}}}\right)\\ =0.989\text{ atm}\end{array}$

The equation for an ideal gas is as follows:

$P{V}_{{\text{NH}}_3}=n_{{\text{NH}}_3}RT$

Rearrange the above equation for the volume of ammonia

$V_{{\text{NH}}_3}=\frac{n_{{\text{NH}}_3}\left(\text{remaining}\right)RT}{P}$

Substitute $2.29\text{ mol}$ for $n$, $287.15\text{ K}$ for $T$, $0.989\text{ atm}$ for $P$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation

$\begin{array}{c}{V}_{{\text{NH}}_{\text{3}}}=\frac{\left(\text{2}\text{.29 }\cancel{\text{mol}}\right)\left(0.08206\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(287.15\cancel{\text{K}}\right)}{\left(0.989\cancel{\text{atm}}\right)}\\ =54.5\text{ L}\end{array}$ $\begin{array}{c}{V}_{{\text{NH}}_{\text{3}}}=\frac{\left(\text{2}\text{.29 mol}\right)\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(287.15\text{ K}\right)}{\left(0.989\text{ atm}\right)}\\ =54.5\text{ L}\end{array}$

${\text{NH}}_4\text{Cl}$ $107\text{ g}$ ${\text{NH}}_3$ $54.5\text{ L}$

Conclusion

Hence, the mass of the ${\text{NH}}_4\text{Cl}$ formed is $107\text{ g}$ and the volume of the gas ${\text{NH}}_3$ left after the reaction is $54.5\text{ L}$.