50g of a good sample of `CaOCl_2`, is made to react with `CO_2`. The volume of `Cl_2` liberated at S.T.P is 

Options

A. 5.6 lit
B. 11.2 lit
C. 22.4 li
D. 4.48 lit

Answer

To solve the problem of determining the volume of chlorine gas (Cl2) liberated when 50g of CaOCl2 reacts with CO2, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between calcium oxychloride (CaOCl2) and carbon dioxide (CO2) can be represented as:
\[ \text{CaOCl}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{Cl}_2 \]
From this reaction, we see that 1 mole of CaOCl2 produces 1 mole of Cl2.

2. **Calculate the Molar Mass of CaOCl2**:
The molar mass of CaOCl2 can be calculated as follows:
- Calcium (Ca): 40.08 g/mol
- Oxygen (O): 16.00 g/mol (2 O atoms = 32.00 g/mol)
- Chlorine (Cl): 35.45 g/mol (2 Cl atoms = 70.90 g/mol)
\[ \text{Molar mass of CaOCl}_2 = 40.08 + 32.00 + 70.90 = 142.98 \, \text{g/mol} \]

3. **Calculate the Number of Moles of CaOCl2**:
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
For 50g of CaOCl2:
\[ \text{Number of moles of CaOCl}_2 = \frac{50 \, \text{g}}{142.98 \, \text{g/mol}} \approx 0.349 \, \text{moles} \]

4. **Determine the Moles of Cl2 Produced**:
From the stoichiometry of the reaction, we know that 1 mole of CaOCl2 produces 1 mole of Cl2. Therefore, the moles of Cl2 produced will also be:
\[ \text{Moles of Cl}_2 = 0.349 \, \text{moles} \]

5. **Calculate the Volume of Cl2 at STP**:
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of Cl2 produced can be calculated as:
\[ \text{Volume of Cl}_2 = \text{Moles of Cl}_2 \times 22.4 \, \text{L/mol} \]
\[ \text{Volume of Cl}_2 = 0.349 \, \text{moles} \times 22.4 \, \text{L/mol} \approx 7.82 \, \text{liters} \]

### Final Answer:
The volume of Cl2 liberated at STP is approximately **7.82 liters**.