What volume of hydrogen sulphide at STP will burn in oxygen to yield `12.8` g of sulphur dioxide according to the equation `2H_(2)S + 3O_(2) to 2H_(2)O + 2SO_(2)`.

Explanation

To solve the problem of finding the volume of hydrogen sulfide (H₂S) that will burn in oxygen to yield 12.8 g of sulfur dioxide (SO₂) at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the combustion of hydrogen sulfide is:
\[ 2 \text{H}_2\text{S} + 3 \text{O}_2 \rightarrow 2 \text{H}_2\text{O} + 2 \text{SO}_2 \]

### Step 2: Calculate the molar mass of SO₂
The molar mass of sulfur dioxide (SO₂) can be calculated as follows:
- Sulfur (S) has a molar mass of approximately 32 g/mol.
- Oxygen (O) has a molar mass of approximately 16 g/mol.
- Therefore, the molar mass of SO₂ is:
\[ 32 \, \text{g/mol} + 2 \times 16 \, \text{g/mol} = 32 + 32 = 64 \, \text{g/mol} \]

### Step 3: Determine the number of moles of SO₂ produced
Using the mass of SO₂ produced (12.8 g), we can calculate the number of moles of SO₂:
\[ \text{Number of moles of SO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.8 \, \text{g}}{64 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 4: Use stoichiometry to find moles of H₂S needed
From the balanced equation, we see that 2 moles of H₂S produce 2 moles of SO₂. Therefore, the moles of H₂S required to produce 0.2 moles of SO₂ is also 0.2 moles:
\[ \text{Moles of H}_2\text{S} = 0.2 \, \text{mol} \]

### Step 5: Calculate the volume of H₂S at STP
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume of H₂S can be calculated as:
\[ \text{Volume of H}_2\text{S} = \text{moles} \times \text{volume per mole} = 0.2 \, \text{mol} \times 22.4 \, \text{L/mol} = 4.48 \, \text{L} \]

### Final Answer
The volume of hydrogen sulfide (H₂S) that will burn in oxygen to yield 12.8 g of sulfur dioxide (SO₂) at STP is **4.48 liters**.

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