At sTP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation
`2H_(2)S(g)+SO_(2)(g)rarr 2H_(2)O(l)+3S(s)`
Which of the following shows that volume of the gas remaining after the reaction?

Options

A. 0.2 litres of `SO_(2)(g)`
B. 0.4 litres of `H_(2)(S)`
C. 1.2 litres of `H_(2)S(g)`
D. 1.2 litres of `SO_(2)(g)`

Answer

To solve the problem, we need to determine the volume of gas remaining after the reaction between hydrogen sulfide (H₂S) and sulfur dioxide (SO₂) at standard temperature and pressure (STP).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The balanced equation for the reaction is:
\[ 2H_2S(g) + SO_2(g) \rightarrow 2H_2O(l) + 3S(s) \]

2. **Convert Volumes to Moles:**
At STP, 1 mole of gas occupies 22.4 liters. We can convert the given volumes of H₂S and SO₂ to moles.
- For H₂S:
\[ \text{Moles of } H_2S = \frac{2.8 \, \text{liters}}{22.4 \, \text{liters/mole}} = 0.125 \, \text{moles} \]
- For SO₂:
\[ \text{Moles of } SO_2 = \frac{1.6 \, \text{liters}}{22.4 \, \text{liters/mole}} = 0.0714 \, \text{moles} \]

3. **Determine the Limiting Reagent:**
The stoichiometry of the reaction shows that 2 moles of H₂S react with 1 mole of SO₂. Therefore, we need to compare the mole ratio.
- Required moles of H₂S for 0.0714 moles of SO₂:
\[ \text{Required moles of } H_2S = 2 \times 0.0714 = 0.1428 \, \text{moles} \]
Since we only have 0.125 moles of H₂S, H₂S is the limiting reagent.

4. **Calculate the Amount of SO₂ Reacted:**
Using the stoichiometry of the reaction:
- For every 2 moles of H₂S, 1 mole of SO₂ is consumed. Therefore, the amount of SO₂ reacted with 0.125 moles of H₂S is:
\[ \text{Moles of } SO_2 \text{ reacted} = \frac{0.125}{2} = 0.0625 \, \text{moles} \]

5. **Calculate Remaining Moles of SO₂:**
- Initial moles of SO₂ = 0.0714 moles
- Moles of SO₂ remaining:
\[ \text{Remaining moles of } SO_2 = 0.0714 - 0.0625 = 0.0089 \, \text{moles} \]

6. **Convert Remaining Moles of SO₂ to Volume:**
To find the volume of the remaining SO₂ at STP:
\[ \text{Volume of remaining } SO_2 = 0.0089 \, \text{moles} \times 22.4 \, \text{liters/mole} = 0.19976 \, \text{liters} \approx 0.2 \, \text{liters} \]

### Final Answer:
The volume of gas remaining after the reaction is approximately **0.2 liters of SO₂**.