In the following reaction, `SO_(2)` acts as a reducing agent:
`SO_(2)+Cl_(2)+2H_(2)O to H_(2)SO_(4)+3HCl`
Find the equivalent weight of `SO_(2)`.

Answer

To find the equivalent weight of \( SO_2 \) in the given reaction, we will follow these steps:

### Step 1: Identify the oxidation states of sulfur in \( SO_2 \) and \( H_2SO_4 \)

1. **In \( SO_2 \)**:
- Let the oxidation state of sulfur be \( x \).
- The oxidation state of oxygen is \(-2\).
- The equation for \( SO_2 \) is:
\[ x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4 \]

2. **In \( H_2SO_4 \)**:
- Let the oxidation state of sulfur be \( y \).
- The equation for \( H_2SO_4 \) is:
\[ y + 2(+1) + 4(-2) = 0 \implies y + 2 - 8 = 0 \implies y - 6 = 0 \implies y = +6 \]

### Step 2: Determine the change in oxidation state

- The change in oxidation state of sulfur from \( SO_2 \) to \( H_2SO_4 \) is:
\[ +6 - (+4) = +2 \]

### Step 3: Calculate the molecular weight of \( SO_2 \)

- The molecular weight of \( SO_2 \) is calculated as follows:
- Atomic weight of sulfur (S) = 32 g/mol
- Atomic weight of oxygen (O) = 16 g/mol
- Therefore, for \( SO_2 \):
\[ \text{Molecular weight of } SO_2 = 32 + 2 \times 16 = 32 + 32 = 64 \text{ g/mol} \]

### Step 4: Determine the valence factor

- The valence factor is equal to the change in oxidation state, which we found to be \( +2 \).

### Step 5: Calculate the equivalent weight of \( SO_2 \)

- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Valence factor}} = \frac{64}{2} = 32 \text{ g/equiv} \]

### Final Answer

The equivalent weight of \( SO_2 \) is **32 g/equiv**.

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