How many moles of `KMnO_(4)` are needed to oxidise a mixture of `1` mole of each `FeSO_(4) & FeC_(2)O_(4)` in acidic medium `:`

Options

A. `(4)/(5)`
B. `(5)/(4)`
C. `(3)/(4)`
D. `(5)/(3)`

Explanation

To determine how many moles of `KMnO4` are needed to oxidize a mixture of `1` mole of each `FeSO4` and `FeC2O4` in acidic medium, we will follow these steps:

### Step 1: Identify the N-factor of `KMnO4`
In acidic medium, the half-reaction for the reduction of `KMnO4` is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
From this equation, we can see that 5 electrons are gained by `KMnO4` during the reduction. Therefore, the N-factor of `KMnO4` is 5.

### Step 2: Determine the N-factor of `FeSO4`
In `FeSO4`, iron is in the +2 oxidation state. When it is oxidized to +3, it loses 1 electron. Thus, the N-factor of `FeSO4` is 1.

### Step 3: Determine the N-factor of `FeC2O4`
In `FeC2O4`, the iron is also in the +2 oxidation state. The oxalate ion `C2O4^2-` can be oxidized to `CO2`, which involves the loss of 2 electrons. Therefore, the total change in oxidation state for `FeC2O4` is:
- 1 electron from iron (from +2 to +3)
- 2 electrons from oxalate (from `C2O4^2-` to `2CO2`)

Thus, the total N-factor for `FeC2O4` is 3 (1 from iron + 2 from oxalate).

### Step 4: Set up the equation for equivalents
The number of equivalents of the oxidizing agent (`KMnO4`) must equal the number of equivalents of the reducing agents (`FeSO4` and `FeC2O4`). The equation can be set up as follows:
\[ \text{Number of moles of } KMnO4 \times N_{\text{KMnO4}} = \text{Number of moles of } FeSO4 \times N_{\text{FeSO4}} + \text{Number of moles of } FeC2O4 \times N_{\text{FeC2O4}} \]

### Step 5: Substitute the values into the equation
Given:
- Number of moles of `FeSO4` = 1
- N-factor of `FeSO4` = 1
- Number of moles of `FeC2O4` = 1
- N-factor of `FeC2O4` = 3

Substituting these values into the equation gives:
\[ \text{Number of moles of } KMnO4 \times 5 = 1 \times 1 + 1 \times 3 \]
\[ \text{Number of moles of } KMnO4 \times 5 = 1 + 3 = 4 \]

### Step 6: Solve for the number of moles of `KMnO4`
\[ \text{Number of moles of } KMnO4 = \frac{4}{5} \]

### Final Answer
Thus, the number of moles of `KMnO4` required to oxidize the mixture is \(\frac{4}{5}\) moles.

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