In order to oxidise a mixture of 1 mole each of FeC_2O_4 ,...

In acidic medium, $KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$. The n-factor for $KMnO_4$ is $5$.

The n-factors for the given reducing agents are calculated based on the total number of moles of electrons lost per mole of the compound:

For $FeC_2O_4$:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}$
Total electrons lost $= 3$, so n-factor $= 3$.
Equivalents $= 1 \times 3 = 3$.

For $Fe_2(C_2O_4)_3$:
$Fe^{3+}$ is not oxidised.
$3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^{-}$
Total electrons lost $= 6$, so n-factor $= 6$.
Equivalents $= 1 \times 6 = 6$.

For $FeSO_4$:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
$SO_4^{2-}$ is not oxidised.
Total electrons lost $= 1$, so n-factor $= 1$.
Equivalents $= 1 \times 1 = 1$.

For $Fe_2(SO_4)_3$:
Neither $Fe^{3+}$ nor $SO_4^{2-}$ can be oxidised further.
n-factor $= 0$.
Equivalents $= 0$.

Total equivalents of the reducing mixture $= 3 + 6 + 1 + 0 = 10$.

Let the number of moles of $KMnO_4$ required be $x$.
Equivalents of $KMnO_4 = x \times 5$.

Equating the equivalents of the oxidising and reducing agents:
$5x = 10$
$x = 2$

Answer: $2$