In order to oxidise a mixture of 1 mole each of $$\text{FeC}_2\text{O}_4$$, $$\text{Fe}_2(\text{C}_2\text{O}_4)_3$$, $$\text{FeSO}_4$$ and $$\text{Fe}_2(\text{SO}_4)_3$$ in acidic medium, the number of moles of $$\text{KMnO}_4$$ required is :

Step 1:

The reduction half-reaction for permanganate in acidic medium:

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

Every mole of $$\text{KMnO}_4$$ therefore accepts $$5$$ electrons.

Step 2:

Now find how many electrons are released when the given mixture is completely oxidised.

Iron centres
Fe is present as $$\text{Fe}^{2+}$$ in $$\text{FeC}_2\text{O}_4$$ and $$\text{FeSO}_4$$, and as $$\text{Fe}^{3+}$$ in $$\text{Fe}_2(\text{C}_2\text{O}_4)_3$$ and $$\text{Fe}_2(\text{SO}_4)_3$$.

• $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$ (loss of one electron)

Number of $$\text{Fe}^{2+}$$ moles = 1 (from $$\text{FeC}_2\text{O}_4$$) + 1 (from $$\text{FeSO}_4$$) = 2

Electrons released by iron = $$2 \times 1 = 2$$

Step 3:

Oxalate ions
In acidic medium $$\text{C}_2\text{O}_4^{2-}$$ is oxidised to $$2\,\text{CO}_2$$:

$$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$$

Each oxalate ion therefore releases 2 electrons.

Moles of oxalate ions present:
 • 1 from $$\text{FeC}_2\text{O}_4$$
 • 3 from $$\text{Fe}_2(\text{C}_2\text{O}_4)_3$$

Total oxalate moles = $$1 + 3 = 4$$

Electrons released by oxalate = $$4 \times 2 = 8$$

Step 4:

Total electrons released by the mixture
$$2 (\text{from Fe}) + 8 (\text{from oxalate}) = 10$$

Moles of $$\text{KMnO}_4$$ required
Since each mole of $$\text{KMnO}_4$$ accepts 5 electrons,

$$\text{Moles of } \text{KMnO}_4 = \frac{10 \text{ electrons}}{5 \text{ electrons per mole}} = 2$$

Hence, the number of moles of $$\text{KMnO}_4$$ needed is 2.

Option B $$=>$$ $$2$$