A `3.0g` sample containing `Fe_(3)O_(4),Fe_(2)O_(3)` and an inert impure substance is treated with excess of `KI` solution in presence of dilute `H_(2)SO_(4)`. The entire iron is converted to `Fe^(2+)` along with the liberation of iodine. The resulting solution is diluted to `100 mL`. A `20 mL` of dilute solution requires `11.0 mL` of `0.5M Na_(2)S_(2)O_(3)` solution to reduce the iodine present. `A` `50 mL` of the diluted solution, after complete extraction of iodine requires `12.80 mL` of `0.25M KMnO_(4)` solution in dilute `H_(2)SO_(4)` medium for the oxidation of `Fe^(2+)`. Calculate the percentage of `Fe_(2)O_(3)` and `Fe_(3)O_(4)` in the original sample.

Options

A. `% Fe_(2)O_(3) = 40`
B. `%FeO = 28`
C. % of inert impurity `=42%`
D. % Inert impurity `=32`

Answer

A, C

Explanation

Let the moles of `Fe_(3)O_(4) = x`
so, moles of `Fe_(2)O_(3)` (in `Fe_(3)O_(4)) = x`.
eq.of `Fe_(2)O_(3)` (in `Fe_(3)O_(4))`
= eq of `KI =eq`.of `I_(2) =`eq of Hypo
(or) `2x = 0.025`
`x = 0.0125`mol of `Fe_(2)O_(3)`
So, mass of `Fe_(2)O_(3) = 0.0125 xx 160 = 2g`
`% Fe_(2)O_(3) = (2)/(5) xx 100 = 40`