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

For, `Fe_(3)O_(4)rarr3FeO, 2e+Fe_(3)^((8//3))rarr3Fe^(2+)`.....(1)
Thus, valence factor for `Fe_(3)O_(4)` is `2` and for `FeO` is `1`.....(2)
Let `"Meq.of" Fe_(3)O_(4)` and `Fe_(2)O_(3)` be `a,b` respectively.
`therefore "Meq.of" Fe_(3)O_(4)+"Meq.of" FeO_(3)`
`="Meq.of" I_(2) "liberated"`
`="Meq.of hypo used"`
`a+b(11xx0.5xx100)/(20)=27.5`.....(3)
Now, the `Fe^(2+)` ions are agin oxidised to `Fe^(3+)` by `KMnO_(4)`. Note that in the change `Fe^(2+)rarrFe^(3+)+e`, valence factor of `Fe^(2+)` is `1`.
Thus,
`"Meq.of" Fe^(2+)( "from" Fe_(3)O_(4))+"Meq.of"Fe^(2+)("from"FeO_(3))`
`="Meq.of" KMnO_(4)"used"`
`[{:("If valence factor for "Fe^(2+)"is" (2)/(3)"from eq."(1)","then,,,,),("Meq.of" Fe^(2+)("from"Fe_(3)O_(4))=a,,,,),("If valence factor for "Fe^(2+)"is"1","then,,,,),("Meq.of" fe^(2+)"(from"Fe_(3)O_(4)=(3a)/(2),,,,),("Similarly,from eq."(2)",Meq.of"Fe^(2+)"(from"Fe_(2)O_(3)=b,,,,):}]`
`:. (3a)/(2)+b=0.25xx5xx12.8xx(100)/(50)=32`
or `3a+2b=64`...(4)
from eqs.(3) and (4),
Meq.of `Fe_(3)=a=9`
`:. w_(Fe_(3)O_(4))=(9xx232)/(2xx1000)=1.044g`
and Meq.of `Fe_(2)O_(3)=b=18.5`
`w_(FeO_(3))=(18.5xx160)/(2xx100)=1.044g`
`:. % "of" fe_(3)O_(4)=(1.044xx100)/(3)=34.8%`
and `% "of" Fe_(2)O_(3)=(1.48xx100)/(4)=49.33%`