How many mole of `FeSO_(4)` reacted with one mole of `KMnO_(4)` in acidic medium?

Options

A. `2/5`
B. 5
C. `1/2`
D. `1/5`

Answer

To determine how many moles of `FeSO4` react with one mole of `KMnO4` in acidic medium, we can follow these steps:

### Step 1: Identify the oxidation states
- In `KMnO4`, manganese (Mn) has an oxidation state of +7.
- In acidic medium, `KMnO4` is reduced to `Mn2+`, where manganese has an oxidation state of +2.

### Step 2: Calculate the change in oxidation state for Mn
- The change in oxidation state for Mn is from +7 to +2.
- Therefore, the change in oxidation state (n-factor) for Mn is:
\[ n = 7 - 2 = 5 \]

### Step 3: Identify the oxidation state of iron in `FeSO4`
- In `FeSO4`, iron (Fe) is in the +2 oxidation state (`Fe^2+`).
- It gets oxidized to `Fe^3+`.

### Step 4: Calculate the change in oxidation state for Fe
- The change in oxidation state for Fe is from +2 to +3.
- Therefore, the n-factor for Fe is:
\[ n = 3 - 2 = 1 \]

### Step 5: Set up the relationship between equivalents
- In a redox reaction, the equivalents of the oxidizing agent must equal the equivalents of the reducing agent.
- The equivalent of `KMnO4` can be calculated as:
\[ \text{Equivalent of } KMnO4 = n \times \text{moles of } KMnO4 = 5 \times 1 = 5 \]
- The equivalent of `FeSO4` can be calculated as:
\[ \text{Equivalent of } FeSO4 = n \times \text{moles of } FeSO4 = 1 \times \text{moles of } FeSO4 \]

### Step 6: Set the equivalents equal
- Setting the equivalents equal gives:
\[ 5 = 1 \times \text{moles of } FeSO4 \]
- Thus, the moles of `FeSO4` that react with 1 mole of `KMnO4` is:
\[ \text{moles of } FeSO4 = 5 \]

### Conclusion
- Therefore, **5 moles of `FeSO4` react with 1 mole of `KMnO4` in acidic medium**.

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