In acidic medium,the required number of moles | vedclass

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Question

(B) In acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$.$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Ferrous oxalat

Vedclass · Accepted Answer

$4.167$

Vedclass · Answer

(B) In acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$.$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$Ferrous oxalate $(FeC_2O_4)$ acts as a reducing agent where $Fe^{2+}$ is oxidized to $Fe^{3+}$ and $C_2O_4^{2-}$ is oxidized to $CO_2$.$FeC_2O_4 ightarrow Fe^{3+} + 2CO_2 + 3e^-$To balance the redox reaction,the number of electrons lost must equal the number of electrons gained.Multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $3$:$3MnO_4^- + 5FeC_2O_4 + 24H^+ ightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$From the balanced equation,$3 \, mol$ of $KMnO_4$ react with $5 \, mol$ of $FeC_2O_4$.Therefore,for $2.5 \, mol$ of $KMnO_4$,the moles of $FeC_2O_4$ required are:$	ext{Moles} = \frac{5}{3} 	imes 2.5 = 4.167 \, mol$.

In acidic medium,the required number of moles of ferrous oxalate $(FeC_2O_4)$ for the reduction of $2.5 \, mol$ of $KMnO_4$ is:

Similar Questions

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