In the following reaction,
`2H_(2)S(g)+SO_(2)(g) rarr 3S(s)+2H_(2)O(l)`
One equivalent of `H_(2)S(g)` will reduce

Options

A. `1` mol `SO_(2)`
B. `0.25` mol `SO_(2)`
C. `0.5` mol `SO_(2)`
D. `2` mol `SO_(2)`

Explanation

To solve the problem, we need to determine how much sulfur dioxide (SO₂) one equivalent of hydrogen sulfide (H₂S) will reduce in the given reaction:

**Reaction:**
\[ 2H_{2}S(g) + SO_{2}(g) \rightarrow 3S(s) + 2H_{2}O(l) \]

### Step 1: Determine the oxidation states
In the reaction:
- The oxidation state of sulfur in H₂S is -2.
- The oxidation state of sulfur in SO₂ is +4.
- The elemental sulfur (S) produced has an oxidation state of 0.

### Step 2: Identify the change in oxidation states
- In H₂S, sulfur changes from -2 to 0 (oxidation).
- In SO₂, sulfur changes from +4 to 0 (reduction).

### Step 3: Calculate the number of equivalents
To calculate the number of equivalents, we need to find the change in oxidation states:
- For H₂S: The change is from -2 to 0, which is a change of +2. Since there are 2 moles of H₂S, the total change for 2 moles is \(2 \times 2 = 4\).
- For SO₂: The change is from +4 to 0, which is a change of -4. The total change for 1 mole of SO₂ is -4.

### Step 4: Determine the equivalent weight
- The equivalent weight of H₂S is calculated as:
\[ \text{Equivalent weight of } H_{2}S = \frac{\text{Molar mass}}{\text{Change in oxidation state}} = \frac{34 \text{ g/mol}}{2} = 17 \text{ g/equiv} \]

- The equivalent weight of SO₂ is calculated as:
\[ \text{Equivalent weight of } SO_{2} = \frac{\text{Molar mass}}{\text{Change in oxidation state}} = \frac{64 \text{ g/mol}}{4} = 16 \text{ g/equiv} \]

### Step 5: Calculate the amount of SO₂ reduced by one equivalent of H₂S
From the balanced reaction, 2 moles of H₂S react with 1 mole of SO₂. Therefore, 1 mole of H₂S will react with:
\[ \text{Amount of SO}_{2} \text{ reduced by 1 equivalent of } H_{2}S = \frac{1 \text{ mole of SO}_{2}}{2 \text{ moles of } H_{2}S} = 0.5 \text{ moles of SO}_{2} \]

### Step 6: Conclusion
Thus, one equivalent of H₂S will reduce **0.25 moles of SO₂**.