Chapter U4.93, Problem 6E

Interpretation Introduction

Interpretation:The percent yield of silver chloride is to be calculated.

Concept introduction: The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 6E

The percent yield of silver chloride is 54.54 %.

Explanation of Solution

The reaction described in Exercise 5is shown below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

The mass of silver nitrate is 6.3 g. The molar mass of silver nitrate is 169.87 g/mol. The number of moles of silver nitrate is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{6.3\text{g}}{169.87\text{g/mol}}\\ =0.0371\text{mol}\end{array}$

The mass of sodium chloride (NaCl) is 4.5 g. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. The number of moles of NaCl is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{4.5\text{g}}{58.44\text{g/mol}}\\ =0.077\text{mol}\end{array}$

In the given reaction, 1.0 mol of silver nitrate reacts with 1.0 mol of sodium chloride. Therefore, 0.077 mol of sodium chloride will react with 0.077 mol of silver nitrate but the available amount is 0.0371 mol of silver nitrate. Therefore, silver nitrate is the limiting reagent.

Now, 1.0 mol of silver nitrate produces 1.0 mol of silver chloride. Therefore, 0.0371 mol of silver nitrate will produce 0.0371 mol of silver chloride.

The molar mass of silver chloride is 143.32 g/mol. The mass (theoretical yield) of silver chloride is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 0.037\text{1}\text{mol}=\frac{\text{Mass}}{143.32\text{g/mol}}\\ \text{Mass}=0.037\text{1}\text{mol}\cdot 143.32\text{g/mol}\\ =\text{5}\text{.317}\text{g}\end{array}$

Therefore, the theoretical yield is 5.317 g. The actual yield is given as 2.9 g. The percent yield is calculated as shown below.

  $\begin{array}{c}\text{Percent}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100\\ =\frac{2.9\text{g}}{5.317\text{g}}\times 100\\ =54.54\%\end{array}$

Therefore, the percent yield of silver chloride is 54.54 %.

Conclusion

The percent yield of silver chloride is 54.54 %.