`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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the moles of sulfur (S).
We know that the mass of sulfur given is 8 grams. To find the number of moles, we use the formula:

\[ \text{Moles of S} = \frac{\text{Given mass}}{\text{Molar mass of S}} \]

The molar mass of sulfur (S) is approximately 32 g/mol.

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

### Step 2: Write the reaction for the formation of sulfur dioxide (SO₂).
When sulfur is burned in the presence of oxygen, it forms sulfur dioxide:

\[ S + O_2 \rightarrow SO_2 \]

### Step 3: Determine the moles of sulfur dioxide (SO₂) produced.
From the balanced equation, we see that 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 4: Oxidation of sulfur dioxide by chlorine water.
The sulfur dioxide (SO₂) can be further oxidized to sulfate ions (SO₄²⁻) in the presence of chlorine water:

\[ SO_2 + Cl_2 \text{ (water)} \rightarrow SO_4^{2-} \]

### Step 5: Determine the moles of sulfate ions (SO₄²⁻) produced.
From the stoichiometry of the reaction, we see that 1 mole of SO₂ produces 1 mole of SO₄²⁻. Therefore, 0.25 moles of SO₂ will produce 0.25 moles of sulfate ions.

### Step 6: Precipitation of barium sulfate (BaSO₄).
When barium chloride (BaCl₂) is added to the solution containing sulfate ions, barium sulfate (BaSO₄) precipitates out:

\[ Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \text{ (s)} \]

### Step 7: Determine the moles of barium sulfate (BaSO₄) formed.
From the stoichiometry of the reaction, we see that 1 mole of sulfate ions produces 1 mole of barium sulfate. Thus, 0.25 moles of sulfate ions will produce 0.25 moles of barium sulfate.

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

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

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