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?

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

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

To solve this problem, we need to determine the volume of KCl solution required to react with a given volume of Pb(NO₃)₂ solution. Let's follow the steps:

1. Write 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]

2. Identify the stoichiometry:
From the balanced equation, 2 moles of KCl react with 1 mole of Pb(NO₃)₂.

3. Calculate the moles of Pb(NO₃)₂:
- Concentration of Pb(NO₃)₂ = 0.210 M
- Volume of Pb(NO₃)₂ solution = 50.0 mL = 0.0500 L

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

4. Calculate the moles of KCl needed:
According to the stoichiometry, we need 2 moles of KCl for every 1 mole of Pb(NO₃)₂:

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

5. Calculate the volume of KCl solution required:
- Concentration of KCl = 0.244 M

[tex]\[
\text{Volume of KCl solution} = \frac{\text{Moles of KCl needed}}{\text{Concentration of KCl}} = \frac{0.0210 \, \text{moles}}{0.244 \, \text{mol/L}} = 0.0861 \, \text{L}
\][/tex]

Convert this volume to mL:

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

Therefore, the volume of 0.244 M KCl solution required is 86.1 mL. Thus, the correct answer is option D) 86.1 mL.