Chapter 15.3, Problem 3RQ

Interpretation Introduction

Interpretation: Mass of $PbC{l}_2$ has to be determined.

Concept introduction: 1 mol of lead nitrate $Pb{(N{O}_3)}_2$ reacts with 2 moles of sodium chloride ( $NaCl$ ) to form 1 mol of lead chloride ( $PbC{l}_2$ ).

The balanced equation: $Pb{(N{O}_3)}_2+2NaCl\to PbC{l}_2+\text{ 2}NaN{O}_3$

Answer to Problem 3RQ

Mass of $PbC{l}_2$ formed is $11.1g.$

Explanation of Solution

Moles of $Pb{(N{O}_3)}_2$ reacted $=(0.200mol\cdot L^{-1})\times (\frac{200.0}{1000}L)$

  $=0.04mol$

Moles of $NaCl$ reacted $=(0.200mol\cdot L^{-1})\times (\frac{200.0}{1000}L)$

  $=0.04mol$

Since both the reactant reacted with equal moles, moles of $PbC{l}_2$ formed $=0.04mol$

Mass of $PbC{l}_2$ formed $=\left(\text{moles of }PbC{l}_2\right)\times \left(\text{molar mass of }PbC{l}_2\right)$

  $=(0.04mol)\times \left(278.1g\cdot mo{l}^{-1}\right)$

  $=11.1g$

Conclusion

Mass of $PbC{l}_2$ formed is $11.1g.$