Chapter 12.3, Problem 38LC

Interpretation Introduction

Interpretation: The percent yield of the given 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 38LC

The percent yield is 70.67%.

Explanation of Solution

The reaction is given as follows:

  $2\text{Al}\left(s\right)+3{\text{CuSO}}_{\text{4}}\left(aq\right)\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+3\text{Cu}\left(s\right)$

The mol of Al is calculated as:

  $\begin{array}{c}\text{Mole}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{1.87\text{g}}{26.98\text{g/mol}}\\ =0.069\text{mol}\end{array}$

According to the balanced equation,

  $\begin{array}{c}2\text{mol}\text{of}\text{Al}\to 3\text{mol}\text{of}\text{Cu}\\ 1\text{mol}\text{of}\text{Al}\to \frac{3}{2}\text{mol}\text{of}\text{Cu}\\ 0.069\text{mol}\text{of}\text{Al}\to \frac{3}{2}\times 0.069\text{mol}\text{of}\text{Cu}\\ \to 0.1035\text{mol}\text{of}\text{Cu}\end{array}$

The mass of copper is calculated as:

  $\begin{array}{c}\text{Mass}=\text{Mole}\times \text{Molar}\text{mass}\\ =0.1035\text{mol}\times 63.546\text{g/mol}\\ =\text{6}\text{.58}\text{g}\end{array}$

The theoretical yield of copper is 6.58 g and the actual yield is given as 4.65 g. The percent yield is calculated as:

  $\begin{array}{c}\text{\%}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100\\ =\frac{4.65\text{g}}{\text{6}\text{.58}\text{g}}\times 100\\ =70.67\%\end{array}$

Hence, the percent yield is 70.67%.