Question

what is the percent yield of the reaction between 5.0 mL of 0.10 M barium chloride and 5.0 mL of 0.10 M silver nitrate if 0.056 grams of the precipitate are produced?

Answer 1

BaCl2 moles = 5 x 0.1 / 1000 = 5 x 10^-4

AgNO3 moles = 5 x 0.1 / 1000 = 5 x 10^-4

BaCl2 + 2 AgNO3 ------------------------> 2 AgCl + Ba(NO3)2

1                 2

5 x 10^-4    5 x 10^-4

limitng reagent is AgNO3

moles of precipitate formed = 5 x 10^-4

mass of precipitate = 5 x 10^-4 x 143.32

                              =0.072 g

percent yield = 0.056 x 100 / 0.072

                       = 78 %