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 the reaction, `KMnO4` (or `MnO4-`) is reduced to `Mn2+` in acidic medium. The oxidation state of manganese in `KMnO4` is +7, and in `Mn2+`, it is +2.

### Step 2: Calculate the change in oxidation state for Mn
The change in oxidation state for manganese is:
\[ +7 \text{ (in KMnO4)} \rightarrow +2 \text{ (in Mn2+)} \]
The change is:
\[ +7 - (+2) = 5 \]
This means that 1 mole of `KMnO4` can accept 5 moles of electrons.

### Step 3: Identify the oxidation of iron
`FeSO4` contains `Fe2+`, which can be oxidized to `Fe3+`. The change in oxidation state for iron is:
\[ +2 \text{ (in Fe2+)} \rightarrow +3 \text{ (in Fe3+)} \]
The change is:
\[ +3 - (+2) = 1 \]
This means that 1 mole of `Fe2+` loses 1 mole of electrons.

### Step 4: Set up the equivalence relationship
In a redox reaction, the number of equivalents of the oxidizing agent must equal the number of equivalents of the reducing agent.

Let \( n \) be the number of moles of `FeSO4` that react with 1 mole of `KMnO4`. The equivalents can be expressed as:
\[ \text{Equivalents of } KMnO4 = n \times \text{valence factor of } Fe2+ \]
\[ \text{Equivalents of } FeSO4 = n \times 1 \]
Since `KMnO4` has a valence factor of 5 (as it accepts 5 electrons), we can write:
\[ 1 \text{ (mole of KMnO4)} \times 5 = n \times 1 \]

### Step 5: Solve for n
From the equation:
\[ 5 = n \]
Thus, \( n = 5 \).

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

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