An aqueous solution containing 5.72 g of lead(II) nitrate is added to an aqueous solution containing 5.85 g
of potassium chloride. The reaction goes to completion, but some was lost in the process of washing and drying the precipitate. The percent yield for the reaction is 81.9%
. How many grams of precipitate is recovered? How many grams of the excess reactant remain? Assume the reaction goes to completion.

3.93g are recovered

Explanation:

Pb(NO3)2 reacts with KCl as follows:

Pb(NO3)2 + 2KCl → 2KNO3 + PbCl2

To solve this question we need to find the moles of each reactant in order to find the limiting reactant:

Moles Pb(NO3)2 -Molar mass: 331.2 g/mol-

5.72g * (1mol/331.2g) = 0.01727 moles

Mole KCl -Molar mass: 74.5513g/mol-

5.85g * (1mol/74.5513g) = 0.07847 moles

For a complete reaction of 0.07847 moles of KCl are required:

0.07847 moles KCl * (1mol Pb(NO3)2 / 2mol KCl) = 0.03923 moles Pb(NO3)2

As there are just 0.01727 moles, Pb(NO3)2 is limiting reactant. Assuming 100% of yield:

Moles PbCl2 = Moles Pb(NO3)2

Mass PbCl2 -Molar mass: 278.1g/mol-

0.01727 moles * (278.1g / mol) = 4.80g

As percent yield is 81.9% = 0.819, the mass of PbCl2 recovered was:

4.80g * 0.819 = 3.93g are recovered