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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.

Answer» <html><body><p></p>Solution :`FeC_2O_4+H_2SO_4toFeSO_4+(COOH)_2` <br/> `Fe^(2+)toFe^(3+)+e^(-)` <br/> `C_2O_4^(2-)to2CO_2+2e^(-)` <br/> 40 " mL of " `(N)/(16)` `KMnO_4=(40)/(16)` m" Eq of "`(Fe^(+2)+OX^(2-))` <br/> Let us <a href="https://interviewquestions.tuteehub.com/tag/take-662846" style="font-weight:bold;" target="_blank" title="Click to know more about TAKE">TAKE</a> y m " mol of "`Fe^(2+)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> m" Eq of "Fe^(+2)=y` (n factor`=1`) <br/> y m " mol of "`OX^(2-)` <br/> `thereforem" Eq of "OX^(2-)=2y` (n factor`=2`) <br/> `thereforey+2y=(40)/(16)implies3y=(40)/(16)becausey=(40)/(48)` <br/> m" Eq of "`Fe^(2+)=(40)/(48)`, `m" Eq of "Fe_2C_2O_4=(40)/(16)` <br/> Second step: <br/> 60 " mL of " `(N)/(16)KMnO_4=(60)/(16)` m" Eq of "total `Fe^(2+)` <a href="https://interviewquestions.tuteehub.com/tag/ltbrgt-2804393" style="font-weight:bold;" target="_blank" title="Click to know more about LTBRGT">LTBRGT</a> `[Fe^(2+)` ions from `FeC_2O_4+Fe^(2+)` ions obtained after the reduction of `Fe_2(SO_4)_3]` <br/> m" Eq of "`Fe_2(SO_4)_3=(60)/(16)-(48)/(48)=(140)/(48)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of <a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a>`(Fe_2(SO_4)_3)/(FeC_2O_4)=(140xx48)/(48xx40)=(7)/(6)=7:6`</body></html>


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