How much amount of `CaCO_3` in gram having percentage purity 25 per cent produces 0.36 litre of` CO_2` at on heating?

Explanation

To solve the problem of how much amount of `CaCO3` with a percentage purity of 25% produces 0.36 liters of `CO2` upon heating, we can follow these steps:

### Step 1: Write the decomposition reaction of Calcium Carbonate
The decomposition of Calcium Carbonate can be represented by the following balanced chemical equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Use Avogadro's Law to relate volume and moles
According to Avogadro's law, one mole of any gas occupies 22.4 liters at standard temperature and pressure (STP).

### Step 3: Calculate moles of `CO2` produced
Given that 0.36 liters of `CO2` is produced, we can calculate the number of moles of `CO2`:
\[ \text{Moles of } CO_2 = \frac{\text{Volume of } CO_2}{22.4 \text{ L/mol}} = \frac{0.36 \text{ L}}{22.4 \text{ L/mol}} \approx 0.01607 \text{ moles} \]

### Step 4: Relate moles of `CaCO3` to moles of `CO2`
From the balanced equation, we see that 1 mole of `CaCO3` produces 1 mole of `CO2`. Therefore, the moles of `CaCO3` required to produce 0.01607 moles of `CO2` is also 0.01607 moles.

### Step 5: Calculate the mass of `CaCO3`
The molar mass of `CaCO3` is approximately 100 g/mol. Thus, the mass of `CaCO3` required is:
\[ \text{Mass of } CaCO_3 = \text{Moles} \times \text{Molar Mass} = 0.01607 \text{ moles} \times 100 \text{ g/mol} \approx 1.607 \text{ g} \]

### Step 6: Account for percentage purity
Since the `CaCO3` has a purity of 25%, we can set up the equation:
\[ \text{Purity} = \frac{\text{Actual mass of } CaCO_3}{\text{Total mass}} \times 100 \]
Let the total mass of `CaCO3` be \( X \) grams. Then:
\[ 25\% = \frac{1.607 \text{ g}}{X} \times 100 \]
This simplifies to:
\[ X = \frac{1.607 \text{ g} \times 100}{25} = 6.428 \text{ g} \]

### Final Answer
The total amount of `CaCO3` required is approximately **6.43 grams**.