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

`Fe_(3)O_(4)` is an equimolar mixture of `Fe_(2)O_(3)` and FeO. Thus, the sample contains `Fe_(2)O_(3), FeO` and impurities. The amount of iodine liberated depends on the amount of `Fe_(2)O_(3)` and the entire iron is converted into `Fe^(2+)`.
`Fe_(3)O_(4)+2KI+H_(2)SO_(4) to 3FeO+H_(2)O+K_(2)SO_(4)+I_(2)`
`Fe_(2)O_(3)+KI+H_(2)SO_(4) to 2FeO+H_(2)O+K_(2)SO_(4)+I_(2)`
`5xx11.0 mL " of " 0.5M Na_(2)S_(2)O_(3)-=55.0 mL " of " 0.5N Na_(2)S_(2)O_(3)` soln.
`-=55.0 mL " of" 0.5N I_(2)` soln.
`-=55.0 mL " of" 0.5N Fe_(2)O_(3)` soln.
`=27.5xx10^(-3) " equivalent" Fe_(3)O_(4)` soln.
`=13.75xx10^(-3) " moles" Fe_(2)O_(3)`
`2xx12.8 mL " of" 0.25 m KMnO_(4)` soln.
`-=25.6 mL " of " 1.25 N KMnO_(4)` soln.
`-=25.6 mL " of" 1.25N FeO` soln.
`=32.0xx10^(-3)` equivalent FeO
`=32.0xx10^(-3)` moles FeO
Moles of FeO in `Fe_(3)O_(4)=0.032-0.0275=0.0045`
Mass of `Fe_(3)O_(4)=0.0045xx232=1.044 g`
Moles of `Fe_(2)O_(3)` existing separately
=0.01375-0.0045=0.00925
Mass of `Fe_(2)O_(3)=0.00925xx160=1.48 g`
`% Fe_(3)O_(4)=(1.044)/(3)xx100=34.8 `
`%Fe_(2)O_(3)=(1.48)/(3)xx100=49.33`