Al + Pb(No3)2 = Al(No3) + Pb 5.00 g sample of aluminum is mixed with a 50.0 g sample of lead (II) nitrate producing 17.12 g of lead. What are the theoretical and percent yields of lead.

Answer 1

First balance the reaction and note my correction to the formula for aluminum nitrate: 2 Al + 3 Pb(NO₃)₂ --> 2 Al(NO₃)₃ + 3 Pb.

For aluminum, 5.00g Al * (1 mol Al /26.98g Al) * (3 mol Pb /2 mol Al) = 0.2780 mol Pb.

For lead (II) nitrate, 50.0g Pb(NO₃)₂ * (1 mol Pb(NO₃)₂ /269.2g Pb(NO₃)₂ ) * (3 mol Pb /3 mol Pb(NO₃)₂ ) = 0.1857 mol Pb. This is the limiting reactant because there would still be Al remaining when the Pb(NO₃)₂ is consumed. Theoretical yield of Pb is 0.1857 mol Pb * (207.2g Pb /1 mol Pb) = 38..48 g Pb. Percent yield is 100 * actual/theoretical = 100 * (17.12/38.48) = 44.49%.