According to the following reaction, what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of [tex]0.210 \, \text{M} \, \text{Pb} \left( \text{NO}_3 \right)_2[/tex] solution?

\[2 \, \text{KCl}(aq) + \text{Pb}\left(\text{NO}_3\right)_2(aq) \rightarrow \text{PbCl}_2(s) + 2 \, \text{KNO}_3(aq)\]

A) 116 mL
B) 97.4 mL
C) 86.1 mL
D) 58.1 mL
E) 43.0 mL

Sure! Let's solve this problem step-by-step using stoichiometry.

1. Find the moles of [tex]$ \text{Pb(NO}_3\text{)}_2 $[/tex]:

You are given a 0.210 M solution of [tex]$ \text{Pb(NO}_3\text{)}_2 $[/tex] and a volume of 50.0 mL. First, convert the volume from mL to L (which is 50.0 mL = 0.0500 L).

[tex]\[
\text{Moles of } \text{Pb(NO}_3\text{)}_2 = \text{Concentration (M)} \times \text{Volume (L)} = 0.210 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.0105 \, \text{mol}
\][/tex]

2. Use the reaction stoichiometry to find the moles of KCl needed:

According to the balanced chemical equation:
[tex]\[
2 \, \text{KCl (aq)} + \text{Pb(NO}_3\text{)}_2 \text{(aq)} \rightarrow \text{PbCl}_2 \text{(s)} + 2 \, \text{KNO}_3 \text{(aq)}
\][/tex]

This reaction indicates that 2 moles of KCl are needed for every 1 mole of [tex]$ \text{Pb(NO}_3\text{)}_2 $[/tex]. So, for 0.0105 moles of [tex]$ \text{Pb(NO}_3\text{)}_2 $[/tex], you need:

[tex]\[
\text{Moles of KCl} = 2 \times 0.0105 \, \text{mol} = 0.0210 \, \text{mol}
\][/tex]

3. Calculate the volume of KCl solution needed:

You have a 0.244 M KCl solution. To find the volume needed to get 0.0210 moles of KCl, use the formula:

[tex]\[
\text{Volume (L)} = \frac{\text{Moles of KCl}}{\text{Concentration (M)}} = \frac{0.0210 \, \text{mol}}{0.244 \, \text{mol/L}} \approx 0.0861 \, \text{L}
\][/tex]

Convert this volume from liters to milliliters:

[tex]\[
0.0861 \, \text{L} \times 1000 = 86.1 \, \text{mL}
\][/tex]

Therefore, the volume of the 0.244 M KCl solution required to react exactly with 50.0 mL of the 0.210 M [tex]$ \text{Pb(NO}_3\text{)}_2 $[/tex] solution is 86.1 mL.

The correct answer is:

C) 86.1 mL