How much amount of `CaCO_3` in gram having percentage purity 25 per cent produces 0.56 litre of` CO_2` at on heating?

Explanation

To solve the problem, we need to determine how much calcium carbonate (CaCO₃) is required to produce 0.56 liters of carbon dioxide (CO₂) when heated, considering that the purity of the CaCO₃ is only 25%.

### Step-by-Step Solution:

1. **Write the Decomposition Reaction:**
The decomposition of calcium carbonate can be represented by the following equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

2. **Use Avogadro's Law:**
According to Avogadro's law, at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, we can relate the volume of CO₂ produced to the moles of CaCO₃ used.

3. **Calculate Moles of CO₂:**
We need to find out how many moles of CO₂ are produced from 0.56 liters:
\[ \text{Moles of CO}_2 = \frac{\text{Volume of CO}_2}{\text{Molar Volume at STP}} = \frac{0.56 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.025 \, \text{moles} \]

4. **Relate Moles of CO₂ to Moles of CaCO₃:**
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CaCO₃ required are also 0.025 moles.

5. **Calculate Mass of Pure CaCO₃:**
The molar mass of CaCO₃ is 100 g/mol. Thus, the mass of pure CaCO₃ needed is:
\[ \text{Mass of pure CaCO}_3 = \text{Moles} \times \text{Molar Mass} = 0.025 \, \text{moles} \times 100 \, \text{g/mole} = 2.5 \, \text{grams} \]

6. **Account for Purity:**
Since the purity of the CaCO₃ is only 25%, we need to find the total mass (X) of the impure CaCO₃ that contains 2.5 grams of pure CaCO₃:
\[ 25\% \text{ of } X = 2.5 \, \text{grams} \]
This can be expressed mathematically as:
\[ \frac{25}{100} \times X = 2.5 \]

7. **Solve for X:**
Rearranging the equation gives:
\[ X = \frac{2.5 \times 100}{25} = 10 \, \text{grams} \]

### Final Answer:
The amount of CaCO₃ required is **10 grams**.