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 g KCl is added to a solution containing 25.6 g Pb²⁺, PbCl₂(s) forms. The solid is filtered and dried and found to have a mass of 29.3 g.
Determine the theoretical yield of PbCl₂.