How much amount of `CaCO_3` in gram having percentage purity 50 per cent produces 0.26 litre of` CO_2` at on heating?

Options

A. 1.16g
B. 2.32g
C. 3.22g
D. 4.4g

Explanation

To solve the problem of how much amount of \( \text{CaCO}_3 \) (calcium carbonate) with a percentage purity of 50% 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 balanced equation:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

### Step 2: Use the ideal gas law to find moles of \( \text{CO}_2 \)
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. We need to find the number of moles of \( \text{CO}_2 \) produced from 0.26 liters:
\[ \text{Moles of } \text{CO}_2 = \frac{\text{Volume of } \text{CO}_2}{22.4 \, \text{L/mol}} = \frac{0.26 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.0116 \, \text{mol} \]

### Step 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 \( \text{CO}_2 \). Therefore, the moles of \( \text{CaCO}_3 \) required to produce 0.0116 moles of \( \text{CO}_2 \) is also 0.0116 moles.

### Step 4: Calculate the mass of \( \text{CaCO}_3 \)
The molar mass of \( \text{CaCO}_3 \) is approximately 100 g/mol. Thus, the mass of \( \text{CaCO}_3 \) needed can be calculated as follows:
\[ \text{Mass of } \text{CaCO}_3 = \text{Moles} \times \text{Molar Mass} = 0.0116 \, \text{mol} \times 100 \, \text{g/mol} = 1.16 \, \text{g} \]

### Step 5: Adjust for percentage purity
Since the purity of the \( \text{CaCO}_3 \) is 50%, we need to find the total mass \( x \) of \( \text{CaCO}_3 \) that contains 1.16 g of pure \( \text{CaCO}_3 \):
\[ 0.50 \times x = 1.16 \, \text{g} \]
Solving for \( x \):
\[ x = \frac{1.16 \, \text{g}}{0.50} = 2.32 \, \text{g} \]

### Final Answer
The amount of \( \text{CaCO}_3 \) required is **2.32 grams**.

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