The volume of `CO_(2)` at STP obtained by heating `1` g of `CaCO_(3)` will be :

Options

A. `1 L`
B. `22.4 L`
C. `0.224 L`
D. `11.2 L`

Answer

To find the volume of CO₂ at STP obtained by heating 1 g of CaCO₃, we can follow these steps:

### Step 1: Determine the molar mass of CaCO₃
The molar mass of calcium carbonate (CaCO₃) can be calculated as follows:
- Calcium (Ca) = 40.08 g/mol
- Carbon (C) = 12.01 g/mol
- Oxygen (O) = 16.00 g/mol (3 atoms of oxygen)

Molar mass of CaCO₃ = 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol

### Step 2: Calculate the number of moles of CaCO₃ in 1 g
To find the number of moles of CaCO₃ in 1 g, use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:
\[ \text{Number of moles of CaCO₃} = \frac{1 \text{ g}}{100.09 \text{ g/mol}} \approx 0.00999 \text{ moles} \]

### Step 3: Determine the moles of CO₂ produced
From the decomposition reaction of CaCO₃:
\[ \text{CaCO₃} \rightarrow \text{CaO} + \text{CO₂} \]
1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will also be approximately 0.00999 moles.

### Step 4: Calculate the volume of CO₂ at STP
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as:
\[ \text{Volume of CO₂} = \text{Number of moles} \times 22.4 \text{ L/mol} \]

Substituting the values:
\[ \text{Volume of CO₂} = 0.00999 \text{ moles} \times 22.4 \text{ L/mol} \approx 0.224 \text{ L} \]

### Final Answer
The volume of CO₂ at STP obtained by heating 1 g of CaCO₃ is approximately **0.224 liters**.

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