[bannanahammock]

Question:

Using .005mol of lead(II) nitrate solution produced 1.902g of lead(II) chloride. What is the percent yield for this reaction?

My reaction was:

3 Pb(NO3)2 + 2 FeCl3 = 3 PbCl2 + 2 Fe(NO3)3

Is this result accurate for yield %?:

.005 mol Pb(NO3)2 [ (3mol PbCl2 / 3mol Pb(NO3)2) ] [ ( 278.0g PbCl2 / mol PbCl2) = 1.39

( 1.902 / 1.39 ) 100 = 136.8%

Doesnt look right to me 

[Borek]

1.39g is a correct mass of PbCl₂ that can be produced from 0.005 mole of Pb(NO₃)₂. 137% doesn't make much sense (unless your product is contaminated), but there is no error in the calculation.