`8 g` of sulphur are burnt to form `SO_(2)`, which is oxidised by `Cl_(2)` water. The solution is treated with `BaCl_(2)` solution. The amount of `BaSO_(4)` precipitated is:

Options

A. 1 mole
B. 0.5 mole
C. 0.24 mole
D. 0.25 mole

Answer

To solve the problem step by step, we will follow the chemical reactions and calculations involved in the process.

### Step 1: Calculate the number of moles of sulfur (S)
We know that the number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular mass}} \]
Given:
- Weight of sulfur = 8 g
- Molecular mass of sulfur (S) = 32 g/mol

Now, substituting the values:
\[ \text{Number of moles of S} = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 2: Write the reaction for the combustion of sulfur
When sulfur is burnt in the presence of oxygen, it forms sulfur dioxide (SO₂):
\[ S + O_2 \rightarrow SO_2 \]

### Step 3: Determine the amount of sulfuric acid (H₂SO₄) produced
The sulfur dioxide (SO₂) can be oxidized in the presence of chlorine (Cl₂) and water to form sulfuric acid (H₂SO₄):
\[ SO_2 + Cl_2 + H_2O \rightarrow H_2SO_4 + 2HCl \]
From the reaction, we see that 1 mole of SO₂ produces 1 mole of H₂SO₄.

Since we have 0.25 moles of SO₂:
\[ 0.25 \text{ moles of SO₂} \rightarrow 0.25 \text{ moles of H₂SO₄} \]

### Step 4: Write the reaction for the formation of barium sulfate (BaSO₄)
When barium chloride (BaCl₂) is added to the solution containing sulfuric acid (H₂SO₄), barium sulfate (BaSO₄) precipitates out:
\[ H_2SO_4 + BaCl_2 \rightarrow BaSO_4 \downarrow + 2HCl \]
From this reaction, we can see that 1 mole of H₂SO₄ produces 1 mole of BaSO₄.

Thus, from 0.25 moles of H₂SO₄, we will get:
\[ 0.25 \text{ moles of H₂SO₄} \rightarrow 0.25 \text{ moles of BaSO₄} \]

### Step 5: Conclusion
The amount of BaSO₄ precipitated is 0.25 moles.

### Final Answer
The amount of BaSO₄ precipitated is **0.25 moles**.

---