The problem asks for the number of moles of $KMnO_4$ required to completely react with one mole of $FeC_2O_4$ in an acidic medium. The possible answers are provided as fractions.

1. Problem Description

The problem asks for the number of moles of $KMnO_4$ required to completely react with one mole of $FeC_2O_4$ in an acidic medium. The possible answers are provided as fractions.

2. Solution Steps

The skeleton equation for the reaction in acidic medium is:

$FeC_2O_4 + KMnO_4 + H^+ \rightarrow Fe^{3+} + CO_2 + Mn^{2+} + H_2O$

First, balance the oxidation half-reaction:

$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$

$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$

Combined: $FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-$

Next, balance the reduction half-reaction:

$MnO_4^- \rightarrow Mn^{2+}$

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Now, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 3 to balance the electrons:

$5FeC_2O_4 \rightarrow 5Fe^{3+} + 10CO_2 + 15e^-$

$3MnO_4^- + 24H^+ + 15e^- \rightarrow 3Mn^{2+} + 12H_2O$

Combine the balanced half-reactions:

$5FeC_2O_4 + 3MnO_4^- + 24H^+ \rightarrow 5Fe^{3+} + 10CO_2 + 3Mn^{2+} + 12H_2O$

From the balanced equation, 5 moles of $FeC_2O_4$ react with 3 moles of $KMnO_4$.

Therefore, 1 mole of $FeC_2O_4$ will react with $\frac{3}{5}$ moles of $KMnO_4$.

3. Final Answer

3/5