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 \, M \, Pb \left( 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) 97.4 mL
B) 116 mL
C) 43.0 mL
D) 86.1 mL

Sure, let's solve this step-by-step.

1. Identify the given information:
- Molarity (concentration) of the [tex]$ Pb(NO_3)_2 $[/tex] solution is 0.210 M.
- Volume of the [tex]$ Pb(NO_3)_2 $[/tex] solution is 50.0 mL.
- Molarity (concentration) of the KCl solution is 0.244 M.

2. Determine the moles of [tex]$ Pb(NO_3)_2 $[/tex] present:
Using the formula for molarity:
[tex]\[
\text{Moles of solute} = \text{Molarity} \times \text{Volume (in liters)}
\][/tex]
Convert 50.0 mL to liters:
[tex]\[
50.0 \text{ mL} = 0.0500 \text{ L}
\][/tex]

Calculate the moles of [tex]$ Pb(NO_3)_2 $[/tex]:
[tex]\[
\text{Moles of } Pb(NO_3)_2 = 0.210 \, \text{M} \times 0.0500 \, \text{L} = 0.0105 \, \text{moles}
\][/tex]

3. Use the stoichiometry of the reaction:
The balanced chemical equation is:
[tex]\[
2 KCl(aq) + Pb(NO_3)_2(aq) \rightarrow PbCl_2(s) + 2 KNO_3(aq)
\][/tex]
According to the equation, 2 moles of KCl react with 1 mole of [tex]$ Pb(NO_3)_2 $[/tex].

Calculate the moles of KCl required:
[tex]\[
\text{Moles of } KCl = 2 \times \text{Moles of } Pb(NO_3)_2 = 2 \times 0.0105 \, \text{moles} = 0.0210 \, \text{moles}
\][/tex]

4. Determine the volume of KCl solution needed:
Using the formula for molarity:
[tex]\[
\text{Volume (in liters)} = \frac{\text{Moles of solute}}{\text{Molarity}}
\][/tex]
Calculate the volume in liters:
[tex]\[
\text{Volume of } KCl = \frac{0.0210 \, \text{moles}}{0.244 \, \text{M}} = 0.0861 \, \text{L}
\][/tex]
Convert the volume to mL:
[tex]\[
0.0861 \, \text{L} = 86.1 \, \text{mL}
\][/tex]

So, the volume of 0.244 M KCl solution required to react exactly with 50.0 mL of 0.210 M [tex]$ Pb(NO_3)_2 $[/tex] solution is 86.1 mL.

Therefore, the correct answer is:
D) 86.1 mL