Question

# Lead ions can be precipitated from solution with KClaccording to the following reaction: Pb2+(aq)+2KCl(aq)→PbCl2(s)+2K+(aq) When 28.8...

Lead ions can be precipitated from solution with KClaccording to the following reaction:
Pb2+(aq)+2KCl(aq)→PbCl2(s)+2K+(aq)
When 28.8 g KCl is added to a solution containing 25.7 gPb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.1 g .

1. Determine the limiting reactant (Pb2+ or KCL)

2. Determine theoretical yield of PbCl2

3. Determine percent yield for the reaction

m = 28.8 g KCl

MW KCl = 74.5513

mol = m/MW = 28.8/74.5513 = 0.386311 mol ofKCl

m = 25.7 g of Pb+2

MW = Pb = 207.2 g/mol

mol = 25.7 /207.2 = 0.12403474903 mol of Pb+2

ratio is 1 mol of Pb : 2 Kcl so

0.386311 mol ofKCl will need 2*0.386311 = 0.772622 mol of Pb+2; we only have 0.12403474903 mol of PB+2 so Pb+2 is limiting the reaction

1)

Pb+2 is the limiting reactant

2)

Theoretical Yield

0.12403474903 mol of Pb+2 must produce 1:1 0.12403474903 mol of PBCl2

mass PbCl2 = 0.12403474903 *278.1 = 34.494 g of PbCl2

3)

% yield = real / theoretical * 100 = 29.1/ 34.494 *100 = 84.3624% yield

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