The volume in litres of `CO_(2)` liberated at STP when 10 grams of 90% pure limestone is heated cmpletely is

Options

A. 22.4 litres
B. 2.24 litres
C. 20.16 litres
D. 2.016 litres

Answer

To solve the problem of finding the volume of carbon dioxide (CO₂) liberated at STP when 10 grams of 90% pure limestone (calcium carbonate, CaCO₃) is heated completely, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Composition of Limestone**:
- Limestone is primarily composed of calcium carbonate (CaCO₃).

2. **Determine the Effective Mass of Calcium Carbonate**:
- Since the limestone is 90% pure, we first calculate the mass of pure CaCO₃ in the 10 grams of limestone.
\[ \text{Mass of pure CaCO}_3 = 10 \, \text{g} \times 0.90 = 9 \, \text{g} \]

3. **Calculate the Number of Moles of Calcium Carbonate**:
- The molar mass of calcium carbonate (CaCO₃) is approximately 100 g/mol.
\[ \text{Number of moles of CaCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{9 \, \text{g}}{100 \, \text{g/mol}} = 0.09 \, \text{moles} \]

4. **Determine the Moles of Carbon Dioxide Produced**:
- From the decomposition reaction:
\[ \text{CaCO}_3 \, \text{(s)} \rightarrow \text{CaO} \, \text{(s)} + \text{CO}_2 \, \text{(g)} \]
- 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.09 moles of CaCO₃ will produce 0.09 moles of CO₂.

5. **Calculate the Volume of Carbon Dioxide at STP**:
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
\[ \text{Volume of CO}_2 = \text{Number of moles} \times 22.4 \, \text{L/mol} = 0.09 \, \text{moles} \times 22.4 \, \text{L/mol} = 2.016 \, \text{L} \]

### Final Answer:
The volume of CO₂ liberated at STP when 10 grams of 90% pure limestone is heated completely is **2.016 liters**.

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