By passing `H_(2)S` gas in acidified `KMnO_(4)` solution, we get

Options

A. `S`
B. `K_(2)S`
C. `MnO_(2)`
D. `K_(2)SO_(3)`

Explanation

To solve the question, we need to analyze the reaction that occurs when hydrogen sulfide (H₂S) gas is passed through an acidified potassium permanganate (KMnO₄) solution.

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are hydrogen sulfide (H₂S) and acidified potassium permanganate (KMnO₄). The KMnO₄ is in an acidic medium, which is important for the reaction.

2. **Understand the Role of KMnO₄**: Potassium permanganate (KMnO₄) is a strong oxidizing agent. In acidic conditions, the MnO₄⁻ ion can be reduced to Mn²⁺.

3. **Write the Half-Reactions**:
- The reduction half-reaction for KMnO₄ in acidic solution is:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
- The oxidation half-reaction for hydrogen sulfide (H₂S) is:
\[ \text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2\text{e}^- \]

4. **Balance the Overall Reaction**: To balance the overall reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
- Multiply the oxidation half-reaction by 5:
\[ 5\text{H}_2\text{S} \rightarrow 5\text{S} + 10\text{H}^+ + 10\text{e}^- \]
- Now, we can combine the two half-reactions:
\[ 2\text{MnO}_4^- + 5\text{H}_2\text{S} + 6\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 5\text{S} + 10\text{H}^+ \]

5. **Identify the Products**: From the balanced reaction, we can see that sulfur (S) is produced as one of the products. The manganese is reduced to Mn²⁺.

6. **Conclusion**: The main product formed when hydrogen sulfide gas is passed through acidified potassium permanganate solution is sulfur (S).

### Final Answer:
The correct answer is **S**.