In order to oxidize a mixture of one mole of each of FeC$$_2$$O$$_4$$, Fe$$_2$$(C$$_2$$O$$_4$$)$$_3$$, FeSO$$_4$$ and Fe$$_2$$(SO$$_4$$)$$_3$$ in acidic medium, the number of moles of KMnO$$_4$$ is:

We have one mole each of $$\mathrm{FeC_2O_4},\; \mathrm{Fe_2(C_2O_4)_3},\; \mathrm{FeSO_4}$$ and $$\mathrm{Fe_2(SO_4)_3}$$. In acidic medium, $$\mathrm{KMnO_4}$$ will oxidise:

1. $$\mathrm{Fe^{2+}\; \to \; Fe^{3+}}$$

2. $$\mathrm{C_2O_4^{2-}\; \to \; 2\,CO_2}$$

The ferric compound $$\mathrm{Fe_2(SO_4)_3}$$ already contains $$\mathrm{Fe^{3+}}$$, so it does not undergo further oxidation.

Oxidation of iron(II)

Each $$\mathrm{Fe^{2+}}$$ loses one electron according to the half-reaction $$\mathrm{Fe^{2+} \to Fe^{3+} + e^{-}}$$.

Number of $$\mathrm{Fe^{2+}}$$ ions present:

$$\mathrm{FeC_2O_4}\; \text{contains one}\; Fe^{2+}$$

$$\mathrm{FeSO_4}\; \text{contains one}\; Fe^{2+}$$

So, total $$\mathrm{Fe^{2+}} = 1 + 1 = 2 \text{ mol}$$

Electrons released:

$$2 \text{ mol Fe}^{2+} \times 1 \dfrac{\text{mol e}^-}{\text{mol Fe}^{2+}} = 2 \text{ mol e}^-$$

Oxidation of oxalate

The half-reaction in acid is

$$\mathrm{C_2O_4^{2-} \to 2\,CO_2 + 2\,e^-}$$

Count of oxalate ions:

$$\mathrm{FeC_2O_4}\; \text{has } 1$$

$$\mathrm{Fe_2(C_2O_4)_3}\; \text{has } 3$$

Total oxalate ions $$= 1 + 3 = 4 \text{ mol}$$

Electrons released:

$$4 \text{ mol C}_2\mathrm{O}_4^{2-} \times 2 \dfrac{\text{mol e}^-}{\text{mol C}_2\mathrm{O}_4^{2-}} = 8 \text{ mol e}^-$$

Total electrons to be accepted

$$2 + 8 = 10 \text{ mol e}^-$$

Reduction of permanganate in acid

The standard half-reaction is

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

So one mole of $$\mathrm{KMnO_4}$$ consumes $$5$$ moles of electrons.

Moles of $$\mathrm{KMnO_4}$$ required

Using the simple proportion

$$\dfrac{10 \text{ mol e}^-}{5 \dfrac{\text{mol e}^-}{\text{mol KMnO}_4}} = 2 \text{ mol KMnO}_4$$

Hence, the correct answer is Option B.