The stoichiometric coefficient of S in the following reaction
`H_(2)S + HNO_(3) to NO + S+ H_(2)O`
is balanced (in acidic medium):

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the stoichiometric coefficient of sulfur (S) in the reaction:

\[ \text{H}_2\text{S} + \text{HNO}_3 \rightarrow \text{NO} + \text{S} + \text{H}_2\text{O} \]

we will balance the equation in acidic medium step by step.

### Step 1: Identify Oxidation States
- In H₂S, sulfur (S) has an oxidation state of -2.
- In the products, sulfur (S) is in the elemental form, which has an oxidation state of 0.
- In HNO₃, nitrogen (N) has an oxidation state of +5, and in NO, nitrogen has an oxidation state of +2.

### Step 2: Write Half-Reactions
**Oxidation Half-Reaction:**
\[ \text{H}_2\text{S} \rightarrow \text{S} \]
- Sulfur is oxidized from -2 to 0, losing 2 electrons.

**Reduction Half-Reaction:**
\[ \text{HNO}_3 \rightarrow \text{NO} \]
- Nitrogen is reduced from +5 to +2, gaining 3 electrons.

### Step 3: Balance the Half-Reactions
**Oxidation Half-Reaction:**
\[ \text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2e^- \]

**Reduction Half-Reaction:**
To balance the reduction half-reaction, we need to add H⁺ ions and water:
\[ \text{HNO}_3 + 3e^- + 3\text{H}^+ \rightarrow \text{NO} + 2\text{H}_2\text{O} \]

### Step 4: Equalize Electrons
To combine the half-reactions, we need to make the number of electrons equal. The oxidation half-reaction loses 2 electrons, and the reduction half-reaction gains 3 electrons. We can multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

**Oxidation (multiplied by 3):**
\[ 3\text{H}_2\text{S} \rightarrow 3\text{S} + 6\text{H}^+ + 6e^- \]

**Reduction (multiplied by 2):**
\[ 2\text{HNO}_3 + 6e^- + 6\text{H}^+ \rightarrow 2\text{NO} + 2\text{H}_2\text{O} \]

### Step 5: Combine the Half-Reactions
Now we can add the balanced half-reactions:
\[ 3\text{H}_2\text{S} + 2\text{HNO}_3 \rightarrow 3\text{S} + 2\text{NO} + 4\text{H}_2\text{O} \]

### Step 6: Final Check
- Count the number of each type of atom on both sides:
- H: 6 on the left (from 3 H₂S) and 8 on the right (4 from 4 H₂O).
- N: 2 on both sides.
- O: 6 on both sides (2 from 2 HNO₃ and 4 from 4 H₂O).
- S: 3 on both sides.

### Conclusion
The stoichiometric coefficient of sulfur (S) in the balanced equation is **3**.