Question

A) Assuming that ferrous ammonium sulfate hexahydrate is the limiting reagent in a reaction with ferrous ammonium sulfate hexahydrate and a potassium iron oxalate complex, compute the theoretical amount of the product formed from 1.00 g Fe (NH4)2 (SO4)2×6H2O.

B) If 1.00 g Fe(NH4)2 (SO4)2×6H2O was used as starting material for the process of ferrous ammonium sulfate hexahydrate and potassium iron oxalate complex, and 1.13 g
of the product K3Fe(C2O4)3×3H2O, what is the percentage yield ofbthe product?

(Please show the work that would be necessary.)

Answer 1

   Fe(NH4)2 (SO4)2×6H2O + K3[Fe(C2O4)3]--------> K3[Fe(C2O4)3] 3H2O + (NH4)2So4 + 3H2O

1mole of Fe(NH4)2 (SO4)2×6H2O gives 1 mole of K3[Fe(C2O4)3] 3H2O

392g of Fe(NH4)2 (SO4)2×6H2O gives 491g of K3[Fe(C2O4)3] 3H2O

1g of Fe(NH4)2 (SO4)2×6H2O gives = 491*1/392   = 1.252g of K3[Fe(C2O4)3] 3H2O

      percentage yield = actual yield*100/theoretical yield

                                   = 1.13*100/1.252 = 90.25%