Chapter 15, Problem 38A

Interpretation Introduction

Interpretation: The number of grams of lead chromate formed when $\text{1}\text{.00}\text{g}$ sample of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is added to $\text{25}\text{.0}\text{ml}$ of $\text{1}\text{.00}\text{M}$ ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ solution is to be calculated.

Concept introduction: The reaction between lead nitrate solution and potassium chromate solution is a double displacement reaction. Lead nitrate and potassium chromate solutions exchange their anions resulting in the formation of lead chromate precipitate and potassium nitrate solution. Unitary method is used to first calculate the limiting reagent. From limiting reagent, the mass of product is found again by using the unitary method. A limiting reagent is the reactant that is present in less amount than required for completing the reaction. It is the one that decides the amount of product formed.

Answer to Problem 38A

The number of grams of lead chromate formed when $\text{1}\text{.00}\text{g}$ sample of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is added to $\text{25}\text{.0}\text{ml}$ of $\text{1}\text{.00}\text{M}$

  ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ solution is $\text{0}\text{.975}\text{g}$ .

Explanation of Solution

The reaction between lead nitrate solution and potassium chromate solution is a double displacement reaction. Here is the balanced reaction:

  ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}{}_{\text{(aq)}}\text{ + Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{}_{\text{(aq)}}\to {\text{ PbCrO}}_{\text{4}}\downarrow {\text{+ 2KNO}}_{\text{3}}{}_{\text{(aq)}}$

It is given that $\text{1}\text{.00}\text{g}$ sample of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is added to $\text{25}\text{.0}\text{ml}$ of $\text{1}\text{.00}\text{M}$

  ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ solution.

Number of moles of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ can be calculated as follows:

  $\begin{array}{c}\mathrm{n}=\text{25}\text{.0ml×1}\text{.00M}\\ =\text{25mmol}\end{array}$

Mass of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ can be calculated as follows:

  $\begin{array}{c}\text{M}={\text{Molar mass of K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{×number of moles}\\ =\text{194}\text{.19}{\text{g/mol×25×10}}^{\text{-3}}\text{mol}\\ =\text{4}\text{.93}\text{g}\end{array}$

Molar mass of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ = $\text{331}\text{.2}\text{g/mol}$

Also,

Molar mass of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ = $\text{194}\text{.19}\text{g/mol}$

From the balanced chemical equation,

  $\text{331}\text{.2}\text{g}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ reacts completely with 194.19 g of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ .

Thus, $\text{1}\text{.00}\text{g}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ will react completely with ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$

  $\begin{array}{c}\mathrm{m}=\frac{194.19}{331.2}\times 1\\ =0.586\text{g}\end{array}$

Available ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\text{4}\text{.93}\text{g}$ .

Thus, ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is present in excess than required for the complete reaction. This also means that $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is the limiting reactant. Its amount decides the amount of lead chromate formed.

Molar mass of ${\text{PbCrO}}_{\text{4}}$ = $\text{323}\text{.19}\text{g/mol}$

From the balanced chemical equation,

  $\text{331}\text{.2}\text{g}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ reacts completely with ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ to form $\text{323}\text{.19}\text{g}$ ${\text{PbCrO}}_{\text{4}}$

So, $\text{1}\text{.00}\text{g}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ will react completely with ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ to form ${\text{PbCrO}}_{\text{4}}$

  $\begin{array}{c}\mathrm{m}=\frac{323.19}{331.2}\times 1\\ =0.975\text{ g}\end{array}$

Conclusion

The number of grams of lead chromate formed when $\text{1}\text{.00}\text{g}$ sample of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is added to $\text{25}\text{.0}\text{ml}$ of $\text{1}\text{.00}\text{M}$ ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ solution is $0.975\text{g}$ .