1. Identify the species that can be oxidised by acidic $KMnO_4$

In acidic medium, $KMnO_4$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ So, $1$ mole of $KMnO_4$ accepts $5$ moles of electrons.

Now check each salt:

• $FeC_2O_4$ contains $Fe^{2+}$ and $C_2O_4^{2-}$
• $Fe_2(C_2O_4)_3$ contains $Fe^{3+}$ and $C_2O_4^{2-}$
• $FeSO_4$ contains $Fe^{2+}$
• $Fe_2(SO_4)_3$ contains $Fe^{3+}$

Only these get oxidised:

• $Fe^{2+} \to Fe^{3+}$
• $C_2O_4^{2-} \to 2CO_2$

$Fe^{3+}$ and $SO_4^{2-}$ are not further oxidised here.

2. Calculate electrons released by each compound

(i) $FeC_2O_4$

It has:

• $1$ mole $Fe^{2+}$ giving $1$ electron: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
• $1$ mole $C_2O_4^{2-}$ giving $2$ electrons: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$

So total electrons released by $1$ mole $FeC_2O_4$: $1 + 2 = 3$

(ii) $Fe_2(C_2O_4)_3$

It has:

• iron already as $Fe^{3+}$, so no oxidation of iron
• $3$ moles of $C_2O_4^{2-}$, each giving $2$ electrons

So total electrons released: $3 \times 2 = 6$

(iii) $FeSO_4$

It has $1$ mole $Fe^{2+}$: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ So electrons released: $1$

(iv) $Fe_2(SO_4)_3$

It contains only $Fe^{3+}$ and $SO_4^{2-}$, so no oxidation occurs.

Electrons released: $0$

3. Total electrons released by the mixture

Since one mole of each salt is present: $3 + 6 + 1 + 0 = 10 \text{ moles of } e^-$

4. Relate electrons to moles of $KMnO_4$

Since $1$ mole of $KMnO_4$ in acidic medium consumes $5$ moles of electrons, $\text{moles of } KMnO_4 = \frac{10}{5} = 2$

5. Answer

$\boxed{2}$ So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.