How many grams of the precipitate are formed?

Question

An aqueous solution containing 9.38 of lead(II) nitrate is added to an aqueous solution containing  of potassium chloride. The percent yield for the reaction is 89.2%. How many grams of the precipitate are formed?

Answer 1

Pb(NO₃)₂(aq) + 2KCl(aq) ==> PbCl₂(s) + 2KNO₃(aq) ... balanced equation

9.38 g Pb(NO₃)₂ x 1 mol Pb(NO₃)₂ / 331 g = 0.0283 mols Pb(NO₃)₂

At 100% yield we would have ...

0.0283 mols Pb(NO₃)₂ x 1 mol PbCl₂ / mol PPb(NO₃)₂ x 70.9 g / mol PbCl2 = 2.00 g PbCl₂

At 89.2% yield, the recovery would be ...

2.00 g x 0.892 = 1.79 g of precipitate PbCl₂