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Answer :
Final answer:
To calculate the volume of KCl solution needed to react with Pb(NO3)2, the number of moles of Pb(NO3)2 is first determined and then the stoichiometric ratios from the balanced equation are used. The required volume of KCl is found by dividing the moles of KCl by its molarity, which is 86.1 mL (option C).
Explanation:
The question relates to stoichiometry and involves reacting potassium chloride (KCl) with lead(II) nitrate (Pb(NO₃)₂) according to the balanced reaction:
2 KCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2 KNO₃(aq)
We first need to calculate the number of moles of Pb(NO₃)₂ present in the 50.0 mL solution with a concentration of 0.210 M.
Moles of Pb(NO₃)₂ = Volume (L) × Molarity (M)
Moles of Pb(NO₃)₂ = 0.050 L × 0.210 M = 0.0105 moles
According to the stoichiometry of the reaction, 1 mole of Pb(NO₃)₂ reacts with 2 moles of KCl. Therefore, to react with 0.0105 moles of Pb(NO₃)₂, we need 0.0210 moles of KCl.
To find the volume of 0.244 M KCl needed we use:
Volume (L) = Moles of KCl / Molarity of KCl
Volume (L) = 0.0210 moles / 0.244 M = 0.0861 L
Since the question provides us with options in milliliters (mL), we need to convert this volume from liters to milliliters.
1 L = 1000 mL
Volume (mL) = 0.0861 L × 1000 mL/L = 86.1 mL
Therefore, the required volume of 0.244 M KCl to react with 50.0 mL of 0.210 M Pb(NO₃)₂ is 86.1 mL, which corresponds to option C).
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