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10 g of a mixture of `Cu_2S` and Cus was titrated with 200 " mL of " 0.75 M `MnO_4^(ɵ)` in acidic medium producing `SO_2,Cu^(2+)`, and `Mn^(2+)`. The `SO_2` was boiled off and the excess of `MnO_4^(ɵ)` was titrated with 175 " mL of " `1 M Fe^(2+)` solution. Find the percentage of CuS the in original mixture. |
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Answer» Total `MnO_4^(ɵ)=200xx0.75M MnO_4^(ɵ)` `=150xx5 mEq` of `MnO_4^(ɵ)` `=750 mEq` of `MnO_4^(ɵ)` Excess of `MnO_4^(ɵ)=175 " mL of " M Fe^(2+)` `=175 m mo`l of `Fe^(2+)` `=175 mEq ` of `Fe^(2+)` `=175 mEq` of ` MnO_4^(ɵ)` `MnO_4^(ɵ)` used up `=750-175=575 mEq` `Cu_2to2Cu^(2+)+2e^(-)` `S^(2-)toSO_2+6e^(-)` 1 " mol of "`Cu_2S` required `8e^(-)` `Ew(Cu_(2)S)=159//8` `CuStoCu^(2+)+S^(2-)` `S^(2-)toSO_(2)+6e^(-)` `therefore` Equivalent weight of `CuS=95.5//6` `therefore` x g of `Cu_2S` and `(10-x)g of CuS` react. `therefore((x)/((159)/(8))+(10-x)/((95.5)/(6)))xx1000=575` `thereforex=` Weight of `Cu_2S=4.25g` `therefore` Weight of `CuS=5.75g` % of `CuS=57.5%` |
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