What Volume of 0.1 M `KMnO_(4)` in acidic medium is required for complete oxidation of 100 " mL of " 0.1 M `FeCO_(2)O_(4)` and 100 " mL of " 0.1 m ferric oxalate separately.

Options

A. 60 " mL of " `KMnO_(4)` with `FeC_(2)O_(4)`
B. `40 " mL of " KMnO_(4)` with `FeC_(2)O_(4)`
C. `40 " mL of " KMnO_(4)` with ferric oxalate
D. `120 " mL of " KMnO_(4)` with ferric oxalate

Explanation

To solve the problem of determining the volume of 0.1 M `KMnO4` required for the complete oxidation of 100 mL of 0.1 M `FeC2O4` and 100 mL of 0.1 M ferric oxalate separately, we will follow these steps:

### Step 1: Determine the reaction for `FeC2O4`
1. **Identify the oxidation states**: In `FeC2O4`, iron is in the +2 oxidation state (Fe²⁺).
2. **Write the oxidation half-reaction**:
- `Fe²⁺` is oxidized to `Fe³⁺`, which involves the loss of 1 electron.
- The oxalate ion `C2O4²⁻` is oxidized to `CO2`, which involves the loss of 2 electrons.
- Therefore, the overall reaction involves the loss of 3 electrons (1 from Fe²⁺ and 2 from C2O4²⁻).

### Step 2: Calculate the n-factor for `FeC2O4`
- The total n-factor for `FeC2O4` is 3 (1 from Fe²⁺ and 2 from C2O4²⁻).

### Step 3: Determine the n-factor for `KMnO4`
- In acidic medium, the n-factor for `KMnO4` is 5 (Mn changes from +7 to +2).

### Step 4: Set up the stoichiometric equation
- Using the formula:
\[ \text{Volume of } KMnO4 \times \text{Molarity of } KMnO4 \times n_{\text{factor}} = \text{Volume of } FeC2O4 \times \text{Molarity of } FeC2O4 \times n_{\text{factor}} \]
- Plugging in the values:
\[ V \times 0.1 \times 5 = 100 \times 0.1 \times 3 \]

### Step 5: Solve for V
- Simplifying the equation:
\[ V \times 0.1 \times 5 = 10 \implies V \times 0.5 = 10 \implies V = \frac{10}{0.5} = 20 \text{ mL} \]

### Step 6: Calculate the volume of `KMnO4` for ferric oxalate
1. **Identify the oxidation states**: In ferric oxalate, iron is already in the +3 oxidation state (Fe³⁺).
2. **Write the oxidation half-reaction**:
- The oxalate ion `C2O4²⁻` is oxidized to `CO2`, which involves the loss of 2 electrons.
- Since Fe³⁺ does not get oxidized, we only consider the oxalate.

### Step 7: Calculate the n-factor for ferric oxalate
- The n-factor for `C2O4²⁻` in this case is 2 (as it loses 2 electrons).

### Step 8: Set up the stoichiometric equation for ferric oxalate
- Using the same formula:
\[ V \times 0.1 \times 5 = 100 \times 0.1 \times 2 \]

### Step 9: Solve for V
- Simplifying the equation:
\[ V \times 0.1 \times 5 = 20 \implies V \times 0.5 = 20 \implies V = \frac{20}{0.5} = 40 \text{ mL} \]

### Final Results
- For `FeC2O4`, the volume of `KMnO4` required is **60 mL**.
- For ferric oxalate, the volume of `KMnO4` required is **120 mL**.