If the theoritical yield of PbCl2 is 16.547 g from the reaction of 7.000 g of NaCl with 50.400 g of Pb(NO3)2. Calculate the % yield of PbCl2 if 12.798 g of PbCl2 was actually obtained? (Atomic mass: Na-22.990 g/mol;0-15.999 g/mol; Cl-35.453 g/mol; Pb-207.200 g/mol; N-14.007).

Answer 1

First, we need to determine the limiting reagent in the reaction. We can do this by calculating the amount of PbCl2 that would be produced if all of the NaCl reacted:

Moles of NaCl = 7.000 g / 58.443 g/mol = 0.1197 mol
Moles of Pb(NO3)2 = 50.400 g / 331.210 g/mol = 0.1522 mol
The reaction between NaCl and Pb(NO3)2 produces one mole of PbCl2 for every two moles of NaCl, so the theoretical yield of PbCl2 from 0.1197 mol of NaCl would be:

0.1197 mol NaCl x (1 mol PbCl2/2 mol NaCl) x 278.109 g/mol = 8.274 g PbCl2

Since the actual yield was 12.798 g, the percent yield is:

Percent yield = (actual yield / theoretical yield) x 100% = (12.798 g / 8.274 g) x 100% = 154.7%

Therefore, the percent yield of PbCl2 is 154.7%.