`SO_(2)Cl_(2)` is dissolved in water to obtain `H_(2)SO_(4)` and HCL. In two different cases, excess `Ba(OH)_(2)` and excess `CH_(3)COOAg` were added to obtain precipitates having masses `m_(1)` and `m_(2)` grams respectively , starting with one mole of `SO_(2)Cl_(2)` in each case. Then (`Ba=137, Ag=108, S=32, Cl=35.5)`:-`

Options

A. `m_(1)=233g`
B. `m_(2)-m_(1)=54g`
C. `m_(2)=287g`
D. All of these

Explanation

To solve the problem, we need to analyze the reactions that occur when `SO₂Cl₂` is dissolved in water and subsequently treated with excess `Ba(OH)₂` and `CH₃COOAg`.

### Step 1: Determine the products of the reaction with water
When `SO₂Cl₂` is dissolved in water, it reacts to form sulfuric acid (`H₂SO₄`) and hydrochloric acid (`HCl`):

\[ \text{SO}_2\text{Cl}_2 + 2 \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + 2 \text{HCl} \]

From this reaction, we see that one mole of `SO₂Cl₂` produces one mole of `H₂SO₄` and two moles of `HCl`.

### Step 2: Reaction with excess `Ba(OH)₂`
Next, we consider the reaction of `H₂SO₄` with excess `Ba(OH)₂`:

\[ \text{H}_2\text{SO}_4 + \text{Ba(OH)}_2 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{H}_2\text{O} \]

Here, `BaSO₄` is formed as a precipitate. Since one mole of `H₂SO₄` is produced, it will react with one mole of `Ba(OH)₂` to produce one mole of `BaSO₄`.

### Step 3: Calculate the mass of `BaSO₄`
The molar mass of `BaSO₄` can be calculated as follows:

- Ba = 137 g/mol
- S = 32 g/mol
- O = 16 g/mol × 4 = 64 g/mol

Total molar mass of `BaSO₄`:

\[ 137 + 32 + 64 = 233 \, \text{g/mol} \]

Since we have one mole of `BaSO₄`, the mass `m₁` is:

\[ m_1 = 1 \, \text{mol} \times 233 \, \text{g/mol} = 233 \, \text{g} \]

### Step 4: Reaction with excess `CH₃COOAg`
Now, we consider the reaction of `HCl` with excess `CH₃COOAg`:

\[ \text{HCl} + \text{CH}_3\text{COOAg} \rightarrow \text{AgCl} \downarrow + \text{CH}_3\text{COOH} \]

Since two moles of `HCl` are produced, it will react with two moles of `CH₃COOAg` to produce two moles of `AgCl`.

### Step 5: Calculate the mass of `AgCl`
The molar mass of `AgCl` can be calculated as follows:

- Ag = 108 g/mol
- Cl = 35.5 g/mol

Total molar mass of `AgCl`:

\[ 108 + 35.5 = 143.5 \, \text{g/mol} \]

Since we have two moles of `AgCl`, the mass `m₂` is:

\[ m_2 = 2 \, \text{mol} \times 143.5 \, \text{g/mol} = 287 \, \text{g} \]

### Step 6: Summary of results
- Mass of `BaSO₄` precipitate, \( m_1 = 233 \, \text{g} \)
- Mass of `AgCl` precipitate, \( m_2 = 287 \, \text{g} \)

### Step 7: Calculate the difference \( m_2 - m_1 \)
Now, we can calculate the difference:

\[ m_2 - m_1 = 287 \, \text{g} - 233 \, \text{g} = 54 \, \text{g} \]

### Final Results
- \( m_1 = 233 \, \text{g} \)
- \( m_2 = 287 \, \text{g} \)
- \( m_2 - m_1 = 54 \, \text{g} \)