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`2Mn_(2)O_(7)rarr4MnO_(2)+3O_(2)` ( if `M` is m ol.wt. of `Mn_(2)O_(7)`). Find the equivalent weight of ` Mn_(2)O_(7)` in above change. |
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Answer» Correct Answer - 6 `Mn_(2)^(7+)+6e rarr2Mn^(4+)` `2xx"mole of" Mn_(2)O_(7)=4xx"mole of" MnO_(2)` `=4xx3xx"eq.of"MnO_(2)` `12 "eq.of" MnO_(2)` `:.` Mole of `Mn_(2)O_(7)=6 "eq.of" MnO_(2)` `because "Eq.of" Mn_(2)O_(7)="Eq.of" MnO_(2)` `:. "Eq.wt.of" Mn_(2)O_(7)=M//6` |
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