Question

What is the experimental yield (in g of precipitate) when 19.6 mL of a 0.6 M solution of iron(III) chloride is combined with 16.4 mL of a 0.662 M solution of silver nitrate at a 75.4% yield?

Answer 1

balanced equation: FeCl3(aq) + 3AgNO3(aq) ----> Fe(NO3)3(aq) + 3AgCl(s)

no of mol of FeCl3 formed = M*V

                          = 19.6*0.6

                          = 11.76 mmol

no of mol of AgNO3 formed = 16.4*0.662

                          = 10.86 mmol

1 mol FeCl3(aq) = 3 mol AgNO3(aq)

no of mol of precipitate(AgCl) = 10.86 mmol

theoretical yield of AgCl formed(T.Y) = n*Mwt = 10.86*10^-3*143.32

                                 = 1.56 g

percent yield of AgCl formed = (P.Y/T.Y)*100

          75.4 = (x/1.56)*100

practical yield of yield of AgCl formed(P.Y) = 1.18 g