In acidic medium,the moles of KMnO_4 required | vedclass

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(D) The oxidation reaction in acidic medium is: $Cu_2S + MnO_4^- \rightarrow Cu^{2+} + SO_2 + Mn^{2+}$.For $Cu_2S$,the oxidation states change as foll

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$2.4$

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(D) The oxidation reaction in acidic medium is: $Cu_2S + MnO_4^- ightarrow Cu^{2+} + SO_2 + Mn^{2+}$.For $Cu_2S$,the oxidation states change as follows: $Cu$ from $+1$ to $+2$ (change of $1$ per $Cu$,total $2$ for $2Cu$) and $S$ from $-2$ to $+4$ (change of $6$).Total $n$-factor for $Cu_2S = 2 + 6 = 8$.For $KMnO_4$,$Mn$ changes from $+7$ to $+2$,so the $n$-factor is $5$.According to the law of equivalence: $	ext{moles of } KMnO_4 	imes (n	ext{-factor of } KMnO_4) = 	ext{moles of } Cu_2S 	imes (n	ext{-factor of } Cu_2S)$.$n 	imes 5 = 1.5 	imes 8$.$n = \frac{12}{5} = 2.4 \, mol$.

In acidic medium,the moles of $KMnO_4$ required to oxidize $1.5 \, mol$ of $Cu_2S$ will be ............... $mol$.

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