Answer :
To solve this problem, we need to figure out how much of the KCl solution is required to fully react with the given amount of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex] solution. Here is a step-by-step breakdown:
1. Identify the reaction and stoichiometry:
The balanced chemical equation is:
[tex]\[
2 \, \text{KCl} \, (aq) + \text{Pb(NO}_3\text{)}_2 \, (aq) \rightarrow \text{PbCl}_2 \, (s) + 2 \, \text{KNO}_3 \, (aq)
\][/tex]
This equation tells us that 2 moles of [tex]\( \text{KCl} \)[/tex] react with 1 mole of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex].
2. Calculate moles of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex]:
- Molarity ([tex]\( M \)[/tex]) of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex] is 0.210 M.
- Volume ([tex]\( V \)[/tex]) of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex] is 50.0 mL or 0.0500 L (since [tex]\( 1 \, \text{L} = 1000 \, \text{mL} \)[/tex]).
Moles of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex]:
[tex]\[
\text{moles} = M \times V = 0.210 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.0105 \, \text{mol}
\][/tex]
3. Determine moles of KCl needed:
According to the stoichiometry of the reaction, 2 moles of [tex]\( \text{KCl} \)[/tex] are needed for every 1 mole of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex]:
[tex]\[
\text{moles of KCl needed} = 2 \times \text{moles of Pb(NO}_3\text{)}_2 = 2 \times 0.0105 \, \text{mol} = 0.0210 \, \text{mol}
\][/tex]
4. Calculate the volume of KCl solution required:
- Molarity ([tex]\( M \)[/tex]) of KCl is 0.244 M.
Volume of KCl solution (in L):
[tex]\[
V = \frac{\text{moles of KCl}}{M} = \frac{0.0210 \, \text{mol}}{0.244 \, \text{mol/L}} = 0.08606557377049182 \, \text{L}
\][/tex]
5. Convert volume from liters to milliliters:
[tex]\[
V \, (\text{in mL}) = 0.08606557377049182 \, \text{L} \times 1000 \, \text{mL/L} = 86.1 \, \text{mL}
\][/tex]
Therefore, the volume of 0.244 M KCl solution required is 86.1 mL. Among the given options, the correct answer is:
c. 86.1 mL
1. Identify the reaction and stoichiometry:
The balanced chemical equation is:
[tex]\[
2 \, \text{KCl} \, (aq) + \text{Pb(NO}_3\text{)}_2 \, (aq) \rightarrow \text{PbCl}_2 \, (s) + 2 \, \text{KNO}_3 \, (aq)
\][/tex]
This equation tells us that 2 moles of [tex]\( \text{KCl} \)[/tex] react with 1 mole of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex].
2. Calculate moles of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex]:
- Molarity ([tex]\( M \)[/tex]) of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex] is 0.210 M.
- Volume ([tex]\( V \)[/tex]) of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex] is 50.0 mL or 0.0500 L (since [tex]\( 1 \, \text{L} = 1000 \, \text{mL} \)[/tex]).
Moles of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex]:
[tex]\[
\text{moles} = M \times V = 0.210 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.0105 \, \text{mol}
\][/tex]
3. Determine moles of KCl needed:
According to the stoichiometry of the reaction, 2 moles of [tex]\( \text{KCl} \)[/tex] are needed for every 1 mole of [tex]\( \text{Pb(NO}_3\text{)}_2 \)[/tex]:
[tex]\[
\text{moles of KCl needed} = 2 \times \text{moles of Pb(NO}_3\text{)}_2 = 2 \times 0.0105 \, \text{mol} = 0.0210 \, \text{mol}
\][/tex]
4. Calculate the volume of KCl solution required:
- Molarity ([tex]\( M \)[/tex]) of KCl is 0.244 M.
Volume of KCl solution (in L):
[tex]\[
V = \frac{\text{moles of KCl}}{M} = \frac{0.0210 \, \text{mol}}{0.244 \, \text{mol/L}} = 0.08606557377049182 \, \text{L}
\][/tex]
5. Convert volume from liters to milliliters:
[tex]\[
V \, (\text{in mL}) = 0.08606557377049182 \, \text{L} \times 1000 \, \text{mL/L} = 86.1 \, \text{mL}
\][/tex]
Therefore, the volume of 0.244 M KCl solution required is 86.1 mL. Among the given options, the correct answer is:
c. 86.1 mL
