Question

Lead(II) ions can be precipitated from solution with KCl according to the following reaction: Pb2+(aq) + 2KCl(aq) → PbCl2(s) + 2K+(aq) When 28.6 gKCl is added to a solution containing 25.8 gPb2+, PbCl2(s) forms. The solid is filtered and dried and found to have a mass of 29.3 g. Determine the limiting reactant.

a. Pb2+ 34.6 g 84.6%
b. KCl 36.2 g 87.3%
c. Pb2+ 25.3 g 78.1%
d. KCl 32.8 g 79.4%

Answer 1

The limiting reactant is Pb2+ because it will be completely consumed first, and the amount of PbCl2 produced from the reaction is determined by the amount of Pb2+ available, not KCl.

Step-by-step explanation:

To identify the limiting reactant in the given reaction, we first need to calculate the moles of each reactant based on their given masses. We will use the molar mass of KCl (74.55 g/mol) and Pb2+ (207.2 g/mol) for this calculation.

For KCl:

28.6 g KCl × (1 mol KCl / 74.55 g KCl) = 0.3837 mol KCl

For Pb2+:

25.8 g Pb2+ × (1 mol Pb2+ / 207.2 g Pb2+) = 0.1245 mol Pb2+

The reaction requires 2 moles of KCl for every mole of Pb2+, hence the molar ratio is 2:1. To determine which reactant is limiting, we compare the number of moles needed with what is available:

• 0.1245 mol Pb2+ requires 0.2490 mol KCl.
• 0.3837 mol KCl is available, which is more than the required amount of KCl.

Since KCl is present in excess, the limiting reactant is Pb2+ because it will be completely consumed first when the reaction goes to completion. As the limiting reactant, Pb2+ will dictate the amount of PbCl2 produced. The precipitated PbCl2 has a mass of 29.3 g, which corresponds to the amount of Pb2+ that reacted.