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.

Explanation

Let the original sample contains x millimol of `Fe_(3)O_(4)` and y millimol of `Fe_(2)O_(3)`. In the first phase of reaction,
`Fe_(3)O_(4) + I^(-) rarr 3FE^(2+) + I_(2)` (n-factor of `Fe_(3)O_(4) =2`)
`Fe_(2)O_(3)+I^(-) rarr 2Fe^(2+) +I_(2)` (n-factor of `Fe_(2)O_(3) = 2`)
`rArr` Meq of `I_(2)` formed = Meq `(Fe_(3)O_(4) +Fe_(2)O_(3))`
= Meq of hydro required
`rArr 2x + 2y = 11 xx 0.5 xx 5 = 27.5` ....(i)
Now, total millimol of `Fe^(2+)` formed `= 3x + 2y`. In the reaction
`Fe^(2+) +MnO_(4)^(-) + H^(+) rarr Fe^(3+) + Mn^(2+)`
n-factor of `Fe^(2+) = 1`
`rArr` Meq of `MnO_(4)^(-) = "Meq of" Fe^(2+)`
`rArr 3x + 2y = 12.8 xx 0.25 xx 5 xx 2 = 32` ...(ii)
solving Eqs. (i) and (ii), we get
`x = 4.5`
and `y = 9.25`
`rArr` Mass of `Fe_(3)O_(4) = (4.5)/(1000) xx 232 = 1.044 g`
`%` mass of `Fe_(3)O_(4) = (1.044)/(3) xx 100 = 34.80%`
Mass of `Fe_(2)O_(3) = (9.25)/(1000) xx 160 = 1.48 g` ltbr. `%` mass of `Fe_(2)O_(3) = (1.48)/(3) xx 100 = 49.33%`