How many moles of KMnO_4 are required to oxid | vedclass

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(A) The balanced half-reaction for the reduction of $MnO_4^-$ in an acidic medium is: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$ This shows that $1 \

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$0.6$

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(A) The balanced half-reaction for the reduction of $MnO_4^-$ in an acidic medium is: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$ This shows that $1 \ mol$ of $KMnO_4$ accepts $5 \ mol$ of electrons. The oxidation of $Fe(C_2O_4)$ involves: $Fe(C_2O_4) 	o Fe^{3+} + 2CO_2 + 3e^-$ This shows that $1 \ mol$ of $Fe(C_2O_4)$ loses $3 \ mol$ of electrons. To balance the electrons,we equate the moles of electrons gained and lost: $5 	imes (	ext{moles of } KMnO_4) = 3 	imes (	ext{moles of } Fe(C_2O_4))$ For $1 \ mol$ of $Fe(C_2O_4)$,the moles of $KMnO_4$ required is: $	ext{Moles of } KMnO_4 = \frac{3}{5} = 0.6 \ mol$.

How many moles of $KMnO_4$ are required to oxidize $1 \ mol$ of $Fe(C_2O_4)$ in an acidic medium?

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