How much amount of `CaCO_3` in gram having percentage purity 25 per cent produces 0.26 litre of` CO_2` at on heating?

Explanation

To solve the problem of how much amount of \( \text{CaCO}_3 \) in grams, having a percentage purity of 25%, produces 0.26 liters of \( \text{CO}_2 \) upon heating, we can follow these steps:

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

### Step 2: Use the molar volume of a gas
According to Avogadro's law, at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters. The molar mass of \( \text{CaCO}_3 \) is 100 g.

### Step 3: Calculate moles of \( \text{CO}_2 \) produced
We need to find out how many moles of \( \text{CO}_2 \) are produced from 0.26 liters:
\[ \text{Moles of } \text{CO}_2 = \frac{\text{Volume of } \text{CO}_2}{\text{Molar volume}} = \frac{0.26 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.0116 \, \text{moles} \]

### Step 4: Relate moles of \( \text{CaCO}_3 \) to moles of \( \text{CO}_2 \)
From the balanced equation, we see that 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). Therefore, the moles of \( \text{CaCO}_3 \) required to produce 0.0116 moles of \( \text{CO}_2 \) is also 0.0116 moles.

### Step 5: Calculate the mass of \( \text{CaCO}_3 \)
Using the molar mass of \( \text{CaCO}_3 \):
\[ \text{Mass of } \text{CaCO}_3 = \text{Moles} \times \text{Molar mass} = 0.0116 \, \text{moles} \times 100 \, \text{g/mole} = 1.16 \, \text{grams} \]

### Step 6: Account for percentage purity
Given that the percentage purity of \( \text{CaCO}_3 \) is 25%, we can set up the equation:
\[ \text{Percentage purity} = \frac{\text{Actual amount of pure } \text{CaCO}_3}{\text{Total amount of } \text{CaCO}_3} \times 100 \]
Let \( X \) be the total amount of \( \text{CaCO}_3 \) needed:
\[ 25\% = \frac{1.16 \, \text{grams}}{X} \times 100 \]
Rearranging gives:
\[ X = \frac{1.16 \, \text{grams} \times 100}{25} = 4.64 \, \text{grams} \]

### Final Answer
The amount of \( \text{CaCO}_3 \) required is **4.64 grams**.