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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.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% |
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Answer» Correct Answer - A,C 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%` |
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