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InterviewSolution
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A sample of `Fe(SO_4)_3` and `FeC_2O_4` was dissolved in `H_2SO_4`. 40 " mL of " `(N)/(16)KMnO_4` was required for complete oxidation. After oxidation the mixture was reduced by `(An)/(H_2SO_4)`. On again oxidation by same `KMnO_4`, 60 mL was required. Calculate the ratio of m" Eq of "`Fe_2(SO_4)_3` and `FeC_2O_4`. |
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Answer» `FeC_2O_4+H_2SO_4toFeSO_4+(COOH)_2` `Fe^(2+)toFe^(3+)+e^(-)` `C_2O_4^(2-)to2CO_2+2e^(-)` 40 " mL of " `(N)/(16)` `KMnO_4=(40)/(16)` m" Eq of "`(Fe^(+2)+OX^(2-))` Let us take y m " mol of "`Fe^(2+)` `therefore m" Eq of "Fe^(+2)=y` (n factor`=1`) y m " mol of "`OX^(2-)` `thereforem" Eq of "OX^(2-)=2y` (n factor`=2`) `thereforey+2y=(40)/(16)implies3y=(40)/(16)becausey=(40)/(48)` m" Eq of "`Fe^(2+)=(40)/(48)`, `m" Eq of "Fe_2C_2O_4=(40)/(16)` Second step: 60 " mL of " `(N)/(16)KMnO_4=(60)/(16)` m" Eq of "total `Fe^(2+)` ltBrgt `[Fe^(2+)` ions from `FeC_2O_4+Fe^(2+)` ions obtained after the reduction of `Fe_2(SO_4)_3]` m" Eq of "`Fe_2(SO_4)_3=(60)/(16)-(48)/(48)=(140)/(48)` Ratio of mEq`(Fe_2(SO_4)_3)/(FeC_2O_4)=(140xx48)/(48xx40)=(7)/(6)=7:6` |
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