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

Explanation

To solve the problem of how many moles of `FeC2O4` (ferrous oxalate) are required to reduce 2 moles of `KMnO4` in acidic medium, we can follow these steps:

### Step 1: Write the Reduction Half-Reaction for KMnO4
The reduction of `KMnO4` in acidic medium can be represented as follows:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
This shows that 1 mole of `MnO4^-` is reduced to `Mn^{2+}` while consuming 5 electrons.

### Step 2: Determine the Total Electrons Needed for 2 Moles of KMnO4
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