A 5g sample containing `FeO_4 (FeO+Fe_2O_3)` and an inert impurity is treated with excess of KI solution in the presence of dilute `H_2SO_4`.The entire Iron converted to Ferrous ion along with liberation of Iodine.The resulting solution is diluted to 100 ml. 20 ml of the diluted solution requires 10 ml of 0.5M `Na_2S_2O_3` solution to reduce the Iodine present.Amongs the following select correct statements.

Options

A. % of `Fe_2O_3` in sample is 40%
B. % of FeO in sample is 28%
C. % of inert impurity in sample is 42%
D. % of inert impurity in sample is 32%

Answer

A,C

Explanation

Let the moles of `Fe_3O_4` is x. So mole of `Fe_2O_3` (in `Fe_3O_4`)=x
eq of `Fe_2O_3` (in `Fe_3O_4`) =eq of KI = eq of `I_2` =eq of `Na_2S_2O_3`
or 2x=0.025
or x=0.0125 mol of `Fe_2O_3`
So mass of `Fe_2O_3=0.0125xx160=2g`
% of `Fe_2O_3=2/5xx100=40%`