`3KClO_(3)+3H_(2)SO_(4)rarr3KHSO_(4)+HClO_(4)+2ClO_(2)+H_(2)O`
Equivalent weight of `KClO_(3)` is

Options

A. `(M)/(4)`
B. `(M)/(2)`
C. `(M+(M)/(2))`
D. `((M)/(4)+(M)/(2))`

Explanation

To find the equivalent weight of \( KClO_3 \) in the given reaction, we will follow these steps:

### Step 1: Write down the balanced chemical equation
The balanced reaction is:
\[ 3KClO_3 + 3H_2SO_4 \rightarrow 3KHSO_4 + HClO_4 + 2ClO_2 + H_2O \]

### Step 2: Identify the type of reaction
This reaction is a disproportionation reaction, where one species is oxidized and reduced simultaneously.

### Step 3: Determine the oxidation states of chlorine
- In \( KClO_3 \), the oxidation state of Cl is +5.
- In \( HClO_4 \), the oxidation state of Cl is +7 (oxidation).
- In \( ClO_2 \), the oxidation state of Cl is +4 (reduction).

### Step 4: Identify the half-reactions
1. **Oxidation half-reaction**:
\[ ClO_3^- \rightarrow ClO_4^- \]
- Change in oxidation state: +5 to +7 (loss of 2 electrons)
- Therefore, \( n_1 = 2 \).

2. **Reduction half-reaction**:
\[ ClO_3^- \rightarrow ClO_2 \]
- Change in oxidation state: +5 to +4 (gain of 1 electron)
- Therefore, \( n_2 = 1 \).

### Step 5: Calculate the equivalent weight
The formula for equivalent weight in a disproportionation reaction is given by:
\[ \text{Equivalent weight} = \frac{M}{n_1 + n_2} \]
Where \( M \) is the molecular weight of \( KClO_3 \), \( n_1 \) is the number of electrons lost in oxidation, and \( n_2 \) is the number of electrons gained in reduction.

### Step 6: Substitute the values
Let \( M \) be the molecular weight of \( KClO_3 \):
\[ \text{Equivalent weight} = \frac{M}{2 + 1} = \frac{M}{3} \]

### Step 7: Conclusion
Thus, the equivalent weight of \( KClO_3 \) can be expressed as \( \frac{M}{3} \).