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.

A `3.0g` sample containing `Fe_(3)O_(4),Fe_(2)O_(3)` and an inert impu

1.	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.
Answer» `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`

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