Chapter 12.3, Problem 33SP

Interpretation Introduction

Interpretation: The percent yield of the reaction is to be stated.

Concept Introduction: According to the law of conservation of mass, the mass can neither be created nor be destroyed. Every reaction must be balanced. With the help of the balanced equation, the limiting reagent can be found and in accordance with it, the amount of product can be calculated.

Answer to Problem 33SP

The percent yield is 57.6%.

Explanation of Solution

The reaction is as follows:

  ${\text{N}}_{\text{2}}+3{\text{H}}_{\text{2}}\to 2{\text{NH}}_{\text{3}}$

The mol of nitrogen and hydrogen is calculated as:

  $\text{Mole}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Substitute the values in the above formula.

  $\begin{array}{c}\text{Mole}\text{of}{\text{N}}_2=\frac{15.0\text{g}}{28\text{g/mol}}\\ =0.536\text{mol}\\ \text{Mole}\text{of}{\text{H}}_2=\frac{15.0\text{g}}{2\text{g/mol}}\\ =7.5\text{mol}\end{array}$

According to the balanced equation,

  $\begin{array}{c}\text{1}\text{mol}\text{of}{\text{N}}_{\text{2}}\to 3\text{mol}\text{of}{\text{H}}_{\text{2}}\\ \text{0}\text{.536}\text{mol}\text{of}{\text{N}}_{\text{2}}\to 3\times 0.536\text{mol}\text{of}{\text{H}}_{\text{2}}\\ \to 1.608\text{mol}\text{of}{\text{H}}_{\text{2}}\end{array}$

Thus, the hydrogen present in excess (7.5 mol is given) so the limiting reagent is nitrogen. The mol of ammonia is calculated as:

  $\begin{array}{c}\text{1}\text{mol}\text{of}{\text{N}}_{\text{2}}\to 2\text{mol}\text{of}{\text{NH}}_{\text{3}}\\ \text{0}\text{.536}\text{mol}\text{of}{\text{N}}_{\text{2}}\to 2\times 0.536\text{mol}\text{of}{\text{NH}}_{\text{3}}\\ \to 1.072\text{mol}\text{of}{\text{NH}}_{\text{3}}\end{array}$

The mass of ammonia is calculated as:

  $\begin{array}{c}\text{Mass}=\text{Mole}\times \text{Molar}\text{mass}\\ =1.072\text{mol}\times 17\text{g/mol}\\ =\text{18}\text{.224}\text{g}\end{array}$

The theoretical yield of copper is 18.224 g and the actual yield is given as 10.5 g. The percent yield is calculated as:

  $\begin{array}{c}\text{\%}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100\\ =\frac{10.5\text{g}}{\text{18}\text{.224}\text{g}}\times 100\\ =57.6\%\end{array}$

Hence, the percent yield is 57.6%.