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Step-by-Step Solution
Step 1: Identify the species and their oxidation behavior in acidic medium
We have a mixture of four compounds, each taken 1 mole:
$\text{FeC}_2\text{O}_4$
$\text{Fe}_2(\text{C}_2\text{O}_4)_3$
$\text{FeSO}_4$
$\text{Fe}_2(\text{SO}_4)_3$
In acidic medium, $\text{KMnO}_4$ typically oxidizes:
$\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ (1 electron lost per each $\text{Fe}^{2+}$),
$\text{C}_2\text{O}_4^{2-}$ (oxalate) to $\text{CO}_2$ (2 electrons lost per 1 mole of oxalate).
For $\text{Fe}_2(\text{SO}_4)_3$, iron is already in the +3 oxidation state and does not undergo further oxidation under these conditions, so it does not contribute to the electron count.
Step 2: Calculate electrons lost by each compound
(a) $\text{FeC}_2\text{O}_4$
Each $\text{Fe}^{2+}$ is oxidized to $\text{Fe}^{3+}$: 1 electron lost.
Each $\text{C}_2\text{O}_4^{2-}$ (oxalate) loses 2 electrons to become 2 $\text{CO}_2$: total 2 electrons lost per oxalate anion.
Therefore, total electrons lost per mole of $\text{FeC}_2\text{O}_4$ = $1 + 2 = 3.$
(b) $\text{Fe}_2(\text{C}_2\text{O}_4)_3$
Iron in $\text{Fe}_2(\text{C}_2\text{O}_4)_3$ is already $\text{Fe}^{3+}$ (so no electron lost there).
There are 3 moles of $\text{C}_2\text{O}_4^{2-}$ per mole of compound, each oxalate loses 2 electrons, so total $3 \times 2 = 6$ electrons lost per mole of $\text{Fe}_2(\text{C}_2\text{O}_4)_3.$
(c) $\text{FeSO}_4$
$\text{Fe}^{2+}$ in $\text{FeSO}_4$ is oxidized to $\text{Fe}^{3+}$: 1 electron lost per mole.
$\text{SO}_4^{2-}$ is not oxidized further in acidic conditions under these circumstances, so no additional electrons lost from the sulfate.
Total electrons lost per mole of $\text{FeSO}_4 = 1.$
(d) $\text{Fe}_2(\text{SO}_4)_3$
Iron is already in the +3 state ($\text{Fe}^{3+}$), so it does not get oxidized. Therefore, 0 electrons are lost.
Step 3: Sum total electrons lost by the mixture
Adding up all the electrons lost for 1 mole of each compound:
$3\text{ (from FeC}_2\text{O}_4) \;+\; 6\text{ (from Fe}_2(\text{C}_2\text{O}_4)_3) \;+\; 1\text{ (from FeSO}_4) \;+\; 0\text{ (from Fe}_2(\text{SO}_4)_3) \;=\; 10.$
Step 4: Relate to moles of $\text{KMnO}_4$ required
In acidic medium, 1 mole of $\text{KMnO}_4$ can accept 5 moles of electrons (from the half-reaction $ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$). Therefore, if $\text{n}$ moles of $\text{KMnO}_4$ are used, the total electrons that can be accepted are $5 \times \text{n}.$
Step 5: Final calculation for $\text{KMnO}_4$
We need to account for a total of 10 electrons lost:
$5 \times \text{n} = 10 \quad \Longrightarrow \quad \text{n} = \dfrac{10}{5} = 2.
Thus, 2 moles of $\text{KMnO}_4$ are required to oxidize the given mixture completely in acidic medium.
Answer: 2
Step-by-Step Solution
Step 1: Identify the species and their oxidation behavior in acidic medium
We have a mixture of four compounds, each taken 1 mole:
$\text{FeC}_2\text{O}_4$
$\text{Fe}_2(\text{C}_2\text{O}_4)_3$
$\text{FeSO}_4$
$\text{Fe}_2(\text{SO}_4)_3$
In acidic medium, $\text{KMnO}_4$ typically oxidizes:
$\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ (1 electron lost per each $\text{Fe}^{2+}$),
$\text{C}_2\text{O}_4^{2-}$ (oxalate) to $\text{CO}_2$ (2 electrons lost per 1 mole of oxalate).
For $\text{Fe}_2(\text{SO}_4)_3$, iron is already in the +3 oxidation state and does not undergo further oxidation under these conditions, so it does not contribute to the electron count.
Step 2: Calculate electrons lost by each compound
(a) $\text{FeC}_2\text{O}_4$
Each $\text{Fe}^{2+}$ is oxidized to $\text{Fe}^{3+}$: 1 electron lost.
Each $\text{C}_2\text{O}_4^{2-}$ (oxalate) loses 2 electrons to become 2 $\text{CO}_2$: total 2 electrons lost per oxalate anion.
Therefore, total electrons lost per mole of $\text{FeC}_2\text{O}_4$ = $1 + 2 = 3.$
(b) $\text{Fe}_2(\text{C}_2\text{O}_4)_3$
Iron in $\text{Fe}_2(\text{C}_2\text{O}_4)_3$ is already $\text{Fe}^{3+}$ (so no electron lost there).
There are 3 moles of $\text{C}_2\text{O}_4^{2-}$ per mole of compound, each oxalate loses 2 electrons, so total $3 \times 2 = 6$ electrons lost per mole of $\text{Fe}_2(\text{C}_2\text{O}_4)_3.$
(c) $\text{FeSO}_4$
$\text{Fe}^{2+}$ in $\text{FeSO}_4$ is oxidized to $\text{Fe}^{3+}$: 1 electron lost per mole.
$\text{SO}_4^{2-}$ is not oxidized further in acidic conditions under these circumstances, so no additional electrons lost from the sulfate.
Total electrons lost per mole of $\text{FeSO}_4 = 1.$
(d) $\text{Fe}_2(\text{SO}_4)_3$
Iron is already in the +3 state ($\text{Fe}^{3+}$), so it does not get oxidized. Therefore, 0 electrons are lost.
Step 3: Sum total electrons lost by the mixture
Adding up all the electrons lost for 1 mole of each compound:
$3\text{ (from FeC}_2\text{O}_4) \;+\; 6\text{ (from Fe}_2(\text{C}_2\text{O}_4)_3) \;+\; 1\text{ (from FeSO}_4) \;+\; 0\text{ (from Fe}_2(\text{SO}_4)_3) \;=\; 10.$
Step 4: Relate to moles of $\text{KMnO}_4$ required
In acidic medium, 1 mole of $\text{KMnO}_4$ can accept 5 moles of electrons (from the half-reaction $ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$). Therefore, if $\text{n}$ moles of $\text{KMnO}_4$ are used, the total electrons that can be accepted are $5 \times \text{n}.$
Step 5: Final calculation for $\text{KMnO}_4$
We need to account for a total of 10 electrons lost:
$5 \times \text{n} = 10 \quad \Longrightarrow \quad \text{n} = \dfrac{10}{5} = 2.
Thus, 2 moles of $\text{KMnO}_4$ are required to oxidize the given mixture completely in acidic medium.
Answer: 2
