How many moles of MnO_4^- are required to com | vedclass

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(C) The balanced redox reaction in acidic medium is: $5FeC_2O_4 + 3MnO_4^- + 24H^+ \rightarrow 5Fe^{3+} + 10CO_2 + 3Mn^{2+} + 12H_2O$ Alternatively,us

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$0.6$

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(C) The balanced redox reaction in acidic medium is: $5FeC_2O_4 + 3MnO_4^- + 24H^+ ightarrow 5Fe^{3+} + 10CO_2 + 3Mn^{2+} + 12H_2O$ Alternatively,using the concept of equivalent weights: $n$-factor of $FeC_2O_4$ (where $Fe^{2+} ightarrow Fe^{3+} + e^-$ and $C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$) is $1 + 2 = 3$. $n$-factor of $MnO_4^-$ in acidic medium is $5$. Equating equivalents: $n_{FeC_2O_4} 	imes (n	ext{-factor}) = n_{MnO_4^-} 	imes (n	ext{-factor})$ $1 	imes 3 = x 	imes 5$ $x = \frac{3}{5} = 0.6 \ mol$.

How many moles of $MnO_4^-$ are required to completely oxidize $1 \ mol$ of ferrous oxalate in an acidic medium?

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