Chapter U4.93, Problem 4E

Interpretation Introduction

Interpretation:The amount of sodium chloride required to produce 50.3 g of mercury (II) nitrate 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 4E

The amount of sodium chloride required to produce 50.3 g of mercury (II) nitrate is 18.12 g.

Explanation of Solution

The reaction described in Exercise 3 is shown below:

  $\text{Hg}\left({\text{NO}}_{\text{3}}\right)\left(aq\right)+2\text{NaCl}\left(aq\right)\to {\text{HgCl}}_{\text{2}}\left(s\right)+2{\text{NaNO}}_{\text{3}}\left(aq\right)$

The mass of mercury (II) nitrate is 50.3 g. The molar mass of mercury (II) nitrate is 324.7 g/mol. The number of moles of mercury (II) nitrate is calculated as shown below:

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\end{array}$

          $\begin{array}{c}=\frac{50.3\text{g}}{324.7\text{g/mol}}\end{array}$

          $\begin{array}{c}=0.155\text{mol}\end{array}$

In the given reaction, 1.0 mol of mercury (II) nitrate reacts with 2.0 mol of sodium chloride. Therefore, the number of moles of sodium chloride required to react with 0.155 mol of mercury (II) nitrate is calculated as shown below:

  $\begin{array}{c}\text{1}\text{.0}\text{mol}\text{of}\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_2\text{reacts}=2.0\text{mol}\text{of}\text{NaCl}\end{array}$

 $\begin{array}{c}\text{0}\text{.155}\text{mol}\text{of}\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_2\text{reacts}=\text{0}\text{.155}\text{mol}\text{of}\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\frac{2.0\text{mol}\text{of}\text{NaCl}}{\text{1}\text{.0}\text{mol}\text{of}\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_2}\right)\end{array}$

              $\begin{array}{c}=0.310\text{mol}\text{of}\text{NaCl}\end{array}$

The molar mass of NaCl is 58.44 g/mol. The mass of NaCl is calculated as shown below:

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\end{array}$

     $\begin{array}{c}0.310\text{mol}=\frac{\text{Mass}}{58.4\text{4}\text{g/mol}}\end{array}$

       $\begin{array}{c}\text{Mass}=0.310\text{mol}\cdot 58.4\text{4}\text{g/mol}\end{array}$

          $\begin{array}{c}=\text{18}\text{.12}\text{g}\end{array}$

Therefore, the mass of NaCl is 18.12 g.

Conclusion

The mass of NaCl is 18.12 g.