When 1 mole of `KMnO_(4)` is reacted with `FeC_(2)O_(4)` in acidic medium, then the reacted amount of `FeC_(2)O_(4)` is

Options

A. `(5)/(3)` moles
B. 5 equivalents
C. `3/5` moles
D. 3 equivalents

Answer

To solve the problem of how much `FeC₂O₄` reacts with 1 mole of `KMnO₄` in acidic medium, we will follow these steps:

### Step 1: Determine the oxidation states
- In `KMnO₄`, manganese (Mn) is in the +7 oxidation state.
- In `FeC₂O₄`, iron (Fe) is in the +2 oxidation state, and carbon (C) in `C₂O₄` has an average oxidation state of +3.

### Step 2: Identify the reduction and oxidation reactions
- `KMnO₄` is reduced from +7 to +2, which means it gains 5 electrons.
- `FeC₂O₄` is oxidized from +2 to +3 for iron and from +3 to +4 for carbon, resulting in a total of 3 electrons being transferred (1 electron for Fe and 2 electrons for the 2 carbons).

### Step 3: Calculate the valency factors
- For `KMnO₄`: The valency factor (n) is 5 (since Mn goes from +7 to +2).
- For `FeC₂O₄`: The valency factor (n) is 3 (1 for Fe and 2 for C).

### Step 4: Set up the equivalent relationship
Using the equation for equivalents:
\[ n_1 \cdot V_1 = n_2 \cdot V_2 \]
Where:
- \( n_1 \) = valency factor of `KMnO₄` = 5
- \( V_1 \) = moles of `KMnO₄` = 1
- \( n_2 \) = valency factor of `FeC₂O₄` = 3
- \( V_2 \) = moles of `FeC₂O₄` (unknown)

### Step 5: Solve for moles of `FeC₂O₄`
Substituting the known values into the equation:
\[ 5 \cdot 1 = 3 \cdot V_2 \]
\[ V_2 = \frac{5}{3} \text{ moles of } FeC₂O₄ \]

### Step 6: Calculate equivalents of `FeC₂O₄`
The equivalents of `FeC₂O₄` can also be calculated:
- Since the valency factor is 3 and we have \( \frac{5}{3} \) moles:
\[ \text{Equivalents of } FeC₂O₄ = n \cdot V_2 = 3 \cdot \frac{5}{3} = 5 \text{ equivalents} \]

### Final Answer
The reacted amount of `FeC₂O₄` is \( \frac{5}{3} \) moles and 5 equivalents.

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