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 will follow these steps:

### Step 1: Determine the number of moles of each reactant.
- The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Volume (L)}}{22.4 \text{ L/mol}} \]
- For Hydrogen Sulfide (H₂S):
\[ \text{Moles of H₂S} = \frac{2.8 \text{ L}}{22.4 \text{ L/mol}} = 0.125 \text{ moles} \]
- For Sulfur Dioxide (SO₂):
\[ \text{Moles of SO₂} = \frac{1.6 \text{ L}}{22.4 \text{ L/mol}} = 0.0714 \text{ moles} \approx 0.07 \text{ moles} \]

### Step 2: Identify the limiting reagent.
- The balanced chemical equation is:
\[ 2 \text{H₂S} + \text{SO₂} \rightarrow 2 \text{H₂O} + 3 \text{S} \]
- According to the stoichiometry of the reaction:
- 2 moles of H₂S react with 1 mole of SO₂.
- To find the limiting reagent, we compare the mole ratio:
- For H₂S:
\[ \frac{0.125 \text{ moles H₂S}}{2} = 0.0625 \]
- For SO₂:
\[ \frac{0.07 \text{ moles SO₂}}{1} = 0.07 \]
- Since 0.0625 (for H₂S) is less than 0.07 (for SO₂), H₂S is the limiting reagent.

### Step 3: Calculate the amount of SO₂ consumed.
- From the balanced equation, 2 moles of H₂S react with 1 mole of SO₂. Therefore, the moles of SO₂ consumed will be:
\[ \text{Moles of SO₂ consumed} = \frac{0.125 \text{ moles H₂S}}{2} = 0.0625 \text{ moles SO₂} \]

### Step 4: Calculate the remaining moles of SO₂.
- Initial moles of SO₂ = 0.07 moles
- Moles of SO₂ remaining after the reaction:
\[ \text{Remaining SO₂} = \text{Initial SO₂} - \text{Consumed SO₂} = 0.07 - 0.0625 = 0.0075 \text{ moles} \]

### Step 5: Convert remaining moles of SO₂ to volume.
- Using the formula:
\[ \text{Volume} = \text{Number of moles} \times 22.4 \text{ L/mol} \]
- Volume of remaining SO₂:
\[ \text{Volume of SO₂} = 0.0075 \text{ moles} \times 22.4 \text{ L/mol} = 0.168 \text{ L} \]

### Conclusion:
The volume of gas remaining after the reaction is approximately **0.168 liters of SO₂**.

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