Excess lead (II) nitrate solution is added to 1.30 g of zinc powder and the mixture is stirred. When the reaction is finished the lead formed is filtered, dried and weighted. It has a mass of 3.31 g. What is the percentage yield of the lead?

Question

Pb(NO3)2 + Zn -> Pb + Zn(NO3)2

A) 80
B) 90
C) 100
D) 70
E) 60

Answer 1

Zn + Pb(NO3)2 ==> Pb + Zn(NO3)2
mols Zn = grams Zn/atomic mas = ?
mols Pb formed = mols Zn
grams Pb = mols Pb x atomic mass Pb. This is the theoretical yield (TY). The actual yield (AY) is given in the problem as 3.31 grams.
% yield = (AY/TY)*100 = ?

Answer 2

Answer is 80% moles of zn =1.30/63=0.02
0.02x207=41
3.31/41x100=80%