Question

A 3.00 g sample containing Fe₃O₄ Fe₂O₃, and an inert impure substance, is treated with excess of KI solution in presence of dilute H₂SO₄. The entire iron is converted into Fe²⁺ along with the liberation of iodine. The resulting solution is diluted to 100 mL. A 20 mL of the diluted solution requires 11.0 mL of 0.5 M Na₂S₂O₃ solution to reduce the iodine, present. A 50 mL of the dilute solution, after complete extraction of the iodine required 12.80 mL of 0.25 M KMnO₄ solution in dilute H₂SO₄ medium for the oxidation of Fe²⁺. Calculate the percentage of Fe₃O₃ and Fe₃O₄ in the original sample.

Answer 1

The following reactions arc involved in given data

No. of millimoles of Na₂S₂O₃ = molarity x volume in mL

= 0.5 x 11 = 5.5

.'. Number of millimoles of I₂ in 20 mL

   = 5.5 x 1/2 = 2.75 

So, the number of millimoles of  I₂ in 100 mL 

= 2.75 x 100/20 = 13.75

No. of millimoles of Fe₂O₃ = 13.75

No. of moles of Fe₂O₃  = 13.75 x 10^-3

Suppose in given sample x moles of Fe₃Oa₄ and y moles of Fe₂O₄ are present. In this sample Fe₂O₄ is an equimolar mixture of FeO and Fe₂O₃. Therefore, total number of moles of Fe₂O₃ in the mixture =  x + y

x + y = 13.75 x 10^-3   ...(i)

No. of miilimoles of KMnO₄ 

= 0.25 x 12.80 = 3.2 

.'. Number of millimoles of FeSO₄ in 50 mL = 3.2 x 5 = 16 

and number of millimoles of FeSO₄ in 100 mL

= 16 x 100/50 = 32

1 mole of Fe₃O₄ gives 3 moles of FeSO₄ and 1 mole of Fe₂O₃ gives 2 moles of FeSO₄

:.  3x + 2y = 32 x 10^-3  ....(ii)

On solving Eqs (i) and (ii)

No. of moles of Fe₃O₄ in the mixture (x)

= 4.5 x 10^-3

No. of moles of Fe₂O₃ in the mixture (y) = 9.25 x 10-3