Question

Lead (II) nitrate and aluminum chloride react according to the following equation: 3Pb(NO3)2 + 2AlCl3 → 3PbCl2 + 2Al(NO3)3. In an experiment, 8.00 g of lead nitrate reacted with 2.67 g of aluminum chloride.

A. Which reactant was the limiting reagent?

B. What is the theoretical yield of lead chloride?

C. If 5.55 g of lead chloride is produced, what is the percent yield (actual amount of product divided by theoretical amount of product)?

Answer 1

3 Pb(NO3)2 + 2 AlCl3 ---------------> 3 PbCl2 + 2 Al(NO3)3

993.6 g            266.68 g                         834.3 g

8.00 g               2.67 g

A)

here limiting reagent is Pb(NO3)2

B)

993.6 g Pb(NO3)2   -----------> 834.3 g PbCl2

8.00 g Pb(NO3)2    ------------->   ??

mass of PbCl2 formed = 8.00 x 834.3 / 993.6

                                     = 6.72 g

theoretical yield of PbCl2 = 6.72 g

C)

percent yield = (actual yield / theoretical yield) x 100

                      = (5.55 / 6.72) x 100

                      = 82.6 %