How many moles of `FeC_2O_4` are required to reduce 2 " mol of "`KMnO_4` in acidic medium?

Explanation

To determine how many moles of `FeC2O4` are required to reduce 2 moles of `KMnO4` in an acidic medium, we need to follow these steps:

### Step 1: Understand the Reaction
In acidic medium, potassium permanganate (`KMnO4`) acts as an oxidizing agent. It gets reduced to manganese(II) ions (`Mn^2+`). The balanced half-reaction for the reduction of `KMnO4` is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 2: Determine the Moles of Electrons
From the balanced half-reaction, we can see that 1 mole of `KMnO4` requires 5 moles of electrons to be reduced. Therefore, for 2 moles of `KMnO4`, the total number of moles of electrons required is:
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