Chapter 9, Problem 4STP

Interpretation Introduction

Interpretation: The mass of ${\text{NH}}_{\text{4}}\text{Cl}$ produced from $10.0\text{ g}$ of ${\text{NH}}_3$ needs to be calculated when ${\text{Cl}}_2$ is present in excess.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

Answer to Problem 4STP

The correct option is (d): $23.6\text{ g}$ .

Explanation of Solution

The given unbalanced reaction is:

  ${\text{NH}}_3{\text{(g) + Cl}}_{\text{2}}\text{(g)}\to {\text{NH}}_{\text{4}}{\text{Cl(s) + NCl}}_3(\text{g)}$

In order to balance the reaction, the coefficient 4 and 3 are written before NH₃ and Cl₂ in the reactant side and coefficient 3 is written before NH₄Cl on the product side. So, the balanced reaction is:

  ${\text{4NH}}_3{\text{(g) + 3Cl}}_{\text{2}}\text{(g)}\to 3{\text{NH}}_{\text{4}}{\text{Cl(s) + NCl}}_3(\text{g)}$

The number of moles of ${\text{NH}}_3$ is calculated as shown:

The molar mass of ${\text{NH}}_3$ is 17.031 g/mol.

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{10\text{ g}}{\text{17}\text{.031 g/mol}}\\ \text{Number of moles}=\text{ 0}\text{.587 mol}\end{array}$

From the balanced chemical reaction, the mole ratio of ${\text{NH}}_3$ and ${\text{NH}}_{\text{4}}\text{Cl}$ is 4:3 so, the number of moles of ${\text{NH}}_{\text{4}}\text{Cl}$ produced from $\text{0}\text{.587 mol}$ of ${\text{NH}}_3$ (as ${\text{Cl}}_2$ is present in excess) is:

  $\text{0}{\text{.587 mol of NH}}_3\text{×}\frac{{\text{3 mol of NH}}_{\text{4}}\text{Cl}}{{\text{4 mol of NH}}_3}\text{ = 0}{\text{.44 mol of NH}}_{\text{4}}\text{Cl}$

Now, the mass of ${\text{NH}}_{\text{4}}\text{Cl}$ produced from $10.0\text{ g}$ of ${\text{NH}}_3$ is calculated as:

The molar mass of ${\text{NH}}_{\text{4}}\text{Cl}$ is 53.491 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.44 mol}=\frac{\text{Mass}}{53.491\text{ g/mol}}\\ \text{Mass = 0}\text{.44 mol}\times 53.491\text{ g/mol}\\ \text{Mass = }23.6\text{ g}\end{array}$

Hence, the mass of ${\text{NH}}_{\text{4}}\text{Cl}$ produced from $10.0\text{ g}$ of ${\text{NH}}_3$ is $23.6\text{ g}$ .