The sulphur dioxide obtained by the combustion of 8 gms of sulphur is passed into Bromine water. The solution is then treated with barium chloride solution. The amount of barium sulphate formed is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of sulfur (S) from the given mass.
The molar mass of sulfur (S) is approximately 32 g/mol.

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

### Step 2: Determine the moles of sulfur dioxide (SO₂) produced.
From the combustion of sulfur, the reaction can be represented as:
\[ S + O_2 \rightarrow SO_2 \]
This means that 1 mole of sulfur produces 1 mole of sulfur dioxide. Therefore, the moles of SO₂ produced will also be:
\[ \text{Moles of SO₂} = 0.25 \text{ moles} \]

### Step 3: React sulfur dioxide with bromine water.
When SO₂ is passed into bromine water, it reacts to form sulfuric acid (H₂SO₄) and hydrobromic acid (HBr):
\[ SO_2 + Br_2 + H_2O \rightarrow H_2SO_4 + 2HBr \]
From this reaction, we see that 1 mole of SO₂ produces 1 mole of H₂SO₄.

### Step 4: Determine the moles of barium sulfate (BaSO₄) formed.
Next, we treat the sulfuric acid with barium chloride (BaCl₂):
\[ H_2SO_4 + BaCl_2 \rightarrow BaSO_4 + 2HCl \]
From this reaction, we see that 1 mole of H₂SO₄ produces 1 mole of BaSO₄. Thus, the moles of BaSO₄ produced will also be:
\[ \text{Moles of BaSO₄} = 0.25 \text{ moles} \]

### Step 5: Calculate the mass of barium sulfate (BaSO₄) formed.
The molar mass of barium sulfate (BaSO₄) can be calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (4 oxygen atoms)

Calculating the total molar mass:
\[ \text{Molar mass of BaSO₄} = 137 + 32 + (16 \times 4) = 137 + 32 + 64 = 233 \text{ g/mol} \]

Now, we can find the mass of BaSO₄ formed:
\[ \text{Mass of BaSO₄} = \text{moles of BaSO₄} \times \text{molar mass of BaSO₄} = 0.25 \text{ moles} \times 233 \text{ g/mol} = 58.25 \text{ g} \]

### Final Answer:
The amount of barium sulfate (BaSO₄) formed is **58.25 grams**.

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