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Answer :
The volume of the 0.244 M KCl solution required to react exactly with 50.0 mL of 0.210 M Pb(NO₃)₂ solution is 86.1 mL.
The correct answer is C) 86.1 mL.
To determine the volume of 0.244 M KCl solution required to react with 50.0 mL of 0.210 M Pb(NO₃)₂ solution, we need to use the stoichiometry of the balanced equation.
The balanced equation is:
2 KCl(aq) +Pb(NO₃)₂ (aq) → PbCl₂(s) + 2 KNO₃(aq)
From the balanced equation, we can see that the molar ratio between KCl and Pb(NO₃)₂ is 2 ratio 1. This means that 2 moles of KCl react with 1 mole of Pb(NO₃)₂ .
First, we need to calculate the number of moles of Pb(NO₃)₂ in the given 50.0 mL of 0.210 M solution:
Moles of Pb(NO₃)₂ = Volume (L) × Concentration (M)
= 0.050 L × 0.210 M
= 0.0105 moles
Since the stoichiometric ratio between Pb(NO₃)₂ and KCl is 1 ratio 2, we can say that 0.0105 moles of Pb(NO₃)₂ will react with 2 × 0.0105 = 0.0210 moles of KCl.
Next, we can calculate the volume of 0.244 M KCl solution needed to provide 0.0210 moles:
Volume (L) = Moles / Concentration
= 0.0210 moles / 0.244 M
= 0.0861 L
Finally, we need to convert the volume to milliliters:
Volume (mL) = 0.0861 L × 1000 mL/L
= 86.1 mL
Therefore, the volume of the 0.244 M KCl solution required to react exactly with 50.0 mL of 0.210 M Pb(NO₃)₂ solution is 86.1 mL.
The correct answer is C) 86.1 mL.
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