1 mol of `SO_(2)` and 1 mol of `H_(2)S` react completely to form `H_(2)O` and S as follows:
`SO_(2)+2H_(2)S to 2H_(2)O+3S`
(At. mass S = 32, O = 16)
The mass of S obtained is:

Options

A. 96 g S
B. 48 g S
C. 24 g S
D. 64 g S

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Identify the limiting reagent
We have 1 mol of SO₂ and 1 mol of H₂S. According to the balanced equation, 1 mol of SO₂ reacts with 2 mol of H₂S. Since we only have 1 mol of H₂S, it will limit the reaction. Therefore, H₂S is the limiting reagent.

### Step 3: Determine the amount of sulfur produced
From the balanced equation, we see that:
- 2 moles of H₂S produce 3 moles of S.

Now, since we have 1 mole of H₂S, we can calculate the moles of sulfur produced:
\[ \text{Moles of S} = \left( \frac{3 \text{ moles of S}}{2 \text{ moles of H}_2\text{S}} \right) \times 1 \text{ mole of H}_2\text{S} = \frac{3}{2} \text{ moles of S} \]

### Step 4: Calculate the mass of sulfur produced
To find the mass of sulfur produced, we use the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
The molar mass of sulfur (S) is 32 g/mol. Therefore:
\[ \text{Mass of S} = \frac{3}{2} \text{ moles of S} \times 32 \text{ g/mol} = 48 \text{ grams} \]

### Conclusion
The mass of sulfur obtained from the reaction is 48 grams.

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