How much amount of `CaCO_3` in gram having percentage purity 25 per cent produces 0.16 litre of` CO_2` at on heating?

Explanation

To solve the problem of how much amount of \( \text{CaCO}_3 \) (calcium carbonate) with a percentage purity of 25% is needed to produce 0.16 liters of \( \text{CO}_2 \) upon heating, we can follow these steps:

### Step-by-Step Solution:

1. **Write the decomposition reaction**:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
This shows that one mole of calcium carbonate decomposes to produce one mole of carbon dioxide.

2. **Use the ideal gas law to relate volume and moles**:
At standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters. Therefore, we can determine how many moles of \( \text{CO}_2 \) are produced from the given volume:
\[ \text{Moles of } \text{CO}_2 = \frac{\text{Volume of } \text{CO}_2}{22.4 \text{ L/mol}} = \frac{0.16 \text{ L}}{22.4 \text{ L/mol}} = 0.00714 \text{ moles} \]

3. **Relate moles of \( \text{CO}_2 \) to moles of \( \text{CaCO}_3 \)**:
From the balanced equation, we see that 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). Thus, the moles of \( \text{CaCO}_3 \) required are also 0.00714 moles.

4. **Calculate the mass of \( \text{CaCO}_3 \)**:
The molar mass of \( \text{CaCO}_3 \) is approximately 100 g/mol. Therefore, the mass of \( \text{CaCO}_3 \) needed can be calculated as follows:
\[ \text{Mass of } \text{CaCO}_3 = \text{Moles} \times \text{Molar mass} = 0.00714 \text{ moles} \times 100 \text{ g/mol} = 0.714 \text{ g} \]

5. **Account for percentage purity**:
Since the purity of the \( \text{CaCO}_3 \) is only 25%, we need to find the total mass \( x \) of the impure \( \text{CaCO}_3 \) that contains 0.714 g of pure \( \text{CaCO}_3 \):
\[ 0.25x = 0.714 \text{ g} \]
Solving for \( x \):
\[ x = \frac{0.714 \text{ g}}{0.25} = 2.856 \text{ g} \]

6. **Round the answer**:
Rounding to two decimal places, we find:
\[ x \approx 2.86 \text{ g} \]

### Final Answer:
The amount of \( \text{CaCO}_3 \) required is approximately **2.86 grams**.