`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 moles `BaSO_(4)` precipitated is:

Explanation

To solve the problem step by step, we will follow the reactions and calculations involved in the process of burning sulfur and forming barium sulfate.

### Step 1: Calculate the moles of sulfur (S)
We start with the mass of sulfur given in the question, which is 8 g. The molar mass of sulfur (S) is approximately 32 g/mol.

\[ \text{Moles of } S = \frac{\text{mass of } S}{\text{molar mass of } S} = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 2: Write the reaction for burning sulfur
When sulfur is burnt in the presence of oxygen, it forms sulfur dioxide (SO₂). The balanced reaction is:

\[ S + O_2 \rightarrow SO_2 \]

From the stoichiometry of this reaction, 1 mole of sulfur produces 1 mole of sulfur dioxide. Therefore, 0.25 moles of sulfur will produce 0.25 moles of sulfur dioxide.

### Step 3: Write the reaction of sulfur dioxide with chlorine water
Next, sulfur dioxide reacts with chlorine and water to form sulfuric acid (H₂SO₄) and hydrochloric acid (HCl). The balanced reaction is:

\[ SO_2 + Cl_2 + H_2O \rightarrow H_2SO_4 + 2HCl \]

From this reaction, we see that 1 mole of sulfur dioxide produces 1 mole of sulfuric acid. Therefore, 0.25 moles of sulfur dioxide will produce 0.25 moles of sulfuric acid.

### Step 4: Write the reaction with barium chloride
When the solution containing sulfuric acid is treated with barium chloride (BaCl₂), barium sulfate (BaSO₄) precipitates out. The balanced reaction is:

\[ H_2SO_4 + BaCl_2 \rightarrow BaSO_4 (s) + 2HCl \]

From this reaction, we see that 1 mole of sulfuric acid produces 1 mole of barium sulfate. Therefore, 0.25 moles of sulfuric acid will produce 0.25 moles of barium sulfate.

### Conclusion
The amount of moles of barium sulfate (BaSO₄) precipitated is:

\[ \text{Moles of } BaSO_4 = 0.25 \text{ moles} \]

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

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