Chapter 11.2, Problem 21SSC

Interpretation Introduction

Interpretation:

The mass of ${\text{NH}}_3$ produced from $2.70\text{ g}$ of ${\text{H}}_2$ needs to be calculated when ${\text{N}}_2$ 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 21SSC

The mass of ${\text{NH}}_3$ produced from $2.70\text{ g}$ of ${\text{H}}_2$ is $15.2\text{ g}$.

Explanation of Solution

The given balanced reaction is:

${\text{N}}_2{\text{(g) + 3H}}_{\text{2}}\text{(g)}\to 2{\text{NH}}_3(\text{g)}$

The number of moles of ${\text{H}}_2$ is calculated as shown:

The molar mass of ${\text{H}}_2$ is 2.016 g/mol.

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{\text{2}\text{.7 g}}{2.016\text{ g/mol}}\\ \text{Number of moles}=\text{ 1}\text{.34 mol}\end{array}$

From the balanced chemical reaction, the mole ratio of ${\text{H}}_2$ and ${\text{NH}}_3$ is 3:2 so, the number of moles of ${\text{NH}}_3$ produced from $\text{1}\text{.34 mol}$ of ${\text{H}}_2$ (as ${\text{N}}_2$ is present in excess) is:

$\text{1}{\text{.34 mol of H}}_2\text{×}\frac{{\text{2 mol of NH}}_3}{{\text{3 mol of H}}_2}\text{ = 0}{\text{.89 mol of NH}}_3$

Now, the mass of ${\text{NH}}_3$ produced from $2.70\text{ g}$ of ${\text{H}}_2$ is calculated as:

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

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.89 mol}=\frac{\text{Mass}}{\text{17}\text{.031 g/mol}}\\ \text{Mass = 0}\text{.89 mol}\times 17.031\text{ g/mol}\\ \text{Mass = 15}\text{.2 g}\end{array}$

Hence, the mass of ${\text{NH}}_3$ produced from $2.70\text{ g}$ of ${\text{H}}_2$ is $15.2\text{ g}$.