Chapter 5, Problem 5.59QP

What is the mass of the solid NH₄Cl formed when 73.0 g of NH₃ are mixed with an equal mass of HCl? What is the volume of the gas remaining, measured at 14.0°C and 752 mmHg? What gas is it?

Interpretation Introduction

Interpretation:

The mass of the solid ${\text{NH}}_{\text{4}}\text{Cl}$ and the volume of the ${\text{NH}}_{\text{3}}$ gas that remains have to be calculated.

Answer to Problem 5.59QP

The volume of ${\text{NH}}_{\text{3}}$ gas is $\text{54}\text{.5L}$.

The mass of ${\text{NH}}_{\text{4}}\text{Cl}$ is $\text{107}\text{g}$.

Explanation of Solution

Chemical equation for the reaction is written as follows:

  ${\text{NH}}_{\text{3(g)}}{\text{ + HCl}}_{\text{(g)}}\to {\text{ NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

To determine the moles of each reactant:

The moles of ${\text{NH}}_{\text{3}}\text{and}\text{HCl}$ gas is calculated by plugging in the values of the given weight of ${\text{NH}}_{\text{3}}\text{and}\text{HCl}$ and molecular weight of the ${\text{NH}}_{\text{3}}$ and $\text{HCl}$ respectively.

  $\begin{array}{c}\text{?mol}{\text{NH}}_{\text{3}}\text{=}\text{73}\text{.0}\text{g}{\text{NH}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{NH}}_{\text{3}}}{\text{17}\text{.03}\text{g}{\text{NH}}_{\text{3}}}\\ \text{=}\text{4}\text{.29}\text{mol}{\text{NH}}_{\text{3}}\\ \text{?mol}\text{HCl}\text{=}\text{73}\text{.0}\text{g}\text{HCl}\text{×}\frac{\text{1}\text{mol}\text{HCl}}{\text{36}\text{.46}\text{g}\text{HCl}}\\ \text{=}\text{2}\text{.00}\text{mol}\text{HCl}\end{array}$

The moles of ${\text{NH}}_{\text{3}}\text{and}\text{HCl}$ gas was found to be $\text{4}\text{.29}\text{mol}$ and $\text{2}\text{.00}\text{mol}$ respectively.

To calculate the mass of ${\text{NH}}_{\text{4}}\text{Cl}$

  $\begin{array}{c}\text{?}\text{g}{\text{NH}}_{\text{4}}\text{Cl}\text{=}\text{2}\text{.00}\text{mol}\text{HCl}\text{×}\frac{\text{1}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}{\text{1}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}\text{×}\frac{\text{53}\text{.49g}{\text{NH}}_{\text{4}}\text{Cl}}{\text{1}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}\\ \text{=}\text{107}\text{g}{\text{NH}}_{\text{4}}\text{Cl}\end{array}$

Given that ${\text{NH}}_{\text{3}}$ and $\text{HCl}$ react in a 1:1 mole ratio, $\text{HCl}$ is the limiting reactant.

The mass of ${\text{NH}}_{\text{4}}\text{Cl}$ is calculated by plugging in the values of the given molecular weight of ${\text{NH}}_{\text{4}}\text{Cl}$ and moles $\text{HCl}$.  The mass of ${\text{NH}}_{\text{4}}\text{Cl}$ was found to be $\text{107}\text{g}{\text{NH}}_{\text{4}}\text{Cl}$

To determine the volume of ${\text{NH}}_{\text{3}}$ using the ideal gas equation.

The gas remaining is ammonia, ${\text{NH}}_{\text{3}}$.  The number of moles of ${\text{NH}}_{\text{3}}$ remaining is $\text{(4}\text{.29-2}\text{.00) mol = 2}\text{.29 mol}\text{.}$   The volume of ${\text{NH}}_{\text{3}}$ gas is calculated by plugging in the values of the given moles, temperature and pressure.

  $\begin{array}{l}\text{ V}\text{=}\frac{\text{nRT}}{\text{P}}\\ {\text{ V}}_{{\text{NH}}_{\text{3}}}\text{=}\frac{\text{(2}\text{.29}\text{mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(14}\text{+}\text{273)}\text{K}}{\left(\text{752}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)}\text{=}\text{54}\text{.5L}{\text{NH}}_{\text{3}}\end{array}$

The volume of ${\text{NH}}_{\text{3}}$ gas was found to be $\text{54}\text{.5L}{\text{NH}}_{\text{3}}$.

Conclusion

The mass of the solid ${\text{NH}}_{\text{4}}\text{Cl}$ and the volume of the ${\text{NH}}_{\text{3}}$ gas was calculated.