How much amount of `CaCO_3` in gram having percentage purity 50 per cent produces 0.36 litre of` CO_2` at on heating?

Options

A. 1.2g
B. 2.2g
C. 3.2g
D. 4.1g

Explanation

To solve the problem, we need to determine how much calcium carbonate (CaCO₃) is required to produce 0.36 liters of carbon dioxide (CO₂) when heated, given that the purity of the CaCO₃ is 50%.

### Step-by-Step Solution:

1. **Write the decomposition reaction of CaCO₃**:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

2. **Determine the molar mass of CaCO₃**:
- The molar mass of CaCO₃ is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
- Total molar mass = 40 + 12 + 48 = 100 g/mol

3. **Use Avogadro's law to find the volume of CO₂ produced**:
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
- From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂.
- Therefore, 100 g of CaCO₃ produces 22.4 liters of CO₂.

4. **Calculate the amount of CaCO₃ needed to produce 0.36 liters of CO₂**:
- Set up a proportion:
\[ \frac{100 \text{ g CaCO}_3}{22.4 \text{ L CO}_2} = \frac{x \text{ g CaCO}_3}{0.36 \text{ L CO}_2} \]
- Cross-multiply to solve for \( x \):
\[ x = \frac{100 \text{ g} \times 0.36 \text{ L}}{22.4 \text{ L}} = \frac{36}{22.4} \approx 1.607 \text{ g} \]

5. **Account for the percentage purity of CaCO₃**:
- Given that the purity is 50%, this means that only 50% of the mass is pure CaCO₃.
- Let \( X \) be the total mass of the impure CaCO₃ required:
\[ 0.50X = 1.607 \text{ g} \]
- Solving for \( X \):
\[ X = \frac{1.607 \text{ g}}{0.50} = 3.214 \text{ g} \]

6. **Final Result**:
- The amount of CaCO₃ required (considering the purity) is approximately 3.21 grams.

### Summary:
The amount of CaCO₃ with 50% purity needed to produce 0.36 liters of CO₂ is approximately **3.21 grams**.