Question

A 25.5-mL sample of a 1.84 M potassium chloride solution is mixed with 14.1 mL of a 0.870 M lead(II) nitrate solution.

2KCL(aq) + Pb(NO₃)₂(aq) -> PbCl₂(s) + 2KNO₃(aq)
The solid PbCl₂ is collected, dried, and found to have a mass of 2.64 grams. Determine the theoretical yield and the percent yield for this reaction.

Answer 1

To determine the theoretical yield of PbCl2 and the percent yield in the reaction between potassium chloride and lead(II) nitrate, we need to calculate the number of moles of each reactant and look at the balanced equation. The limiting reactant is the one that produces the least amount of product, and the theoretical yield is the maximum amount of product that can be formed.

Step-by-step explanation:

In this reaction, the 1.84 M potassium chloride solution and the 0.870 M lead(II) nitrate solution react to form solid lead(II) chloride (PbCl₂) and potassium nitrate (KNO₃) in solution. To determine the theoretical yield of PbCl₂, we need to calculate the number of moles of each reactant and look at the balanced equation. The limiting reactant is the one that produces the least amount of product, and the theoretical yield is the maximum amount of product that can be formed. The percent yield is calculated by dividing the actual yield (2.64 grams) by the theoretical yield and multiplying by 100.