Question

What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction. (Hint Use unit analysis)

2 KCl(aq) + Pb(NO₃)2(aq) → PbCl₂(s) + 2 KNO₃(aq)?

Answer 1

The volume of 0.150 M KCl required to fully react with 0.150 L of 0.175 M Pb(NO3)2 is 350 mL. The calculation is based on the stoichiometry of the reaction, which demands a 2:1 ratio of moles of KCl to Pb(NO3)2.

Step-by-step explanation:

The student has asked for the volume of 0.150 M KCl solution required to completely react with 0.150 L of 0.175 M Pb(NO3)2. According to the balanced chemical equation given in the question, 2 moles of KCl react with 1 mole of Pb(NO3)2. Therefore, the stoichiometry of the reaction is 2:1.

To find the volume of KCl needed, we calculate the moles of Pb(NO3)2 present in 0.150 L of the solution:

Moles of Pb(NO3)2 = Molarity of Pb(NO3)2 × Volume of Pb(NO3)2 solution

Moles of Pb(NO3)2 = 0.175 mol/L × 0.150 L = 0.02625 mol

According to the stoichiometry, we need twice as many moles of KCl. Thus:

Moles of KCl needed = 2 × Moles of Pb(NO3)2 = 2 × 0.02625 mol = 0.0525 mol

To find the volume of the KCl solution, we use the molarity of KCl:

Volume of KCl = Moles of KCl needed ÷ Molarity of KCl = 0.0525 mol ÷ 0.150 mol/L = 0.35 L or 350 mL

Therefore, 350 mL of 0.150 M KCl solution is required to completely react with 0.150 L of 0.175 M Pb(NO3)2.