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 solve the problem of finding the volume of \( CO_2 \) at STP obtained by heating 1 g of \( CaCO_3 \), we will follow these steps:

### Step 1: Write the decomposition reaction of \( CaCO_3 \)
When calcium carbonate (\( CaCO_3 \)) is heated, it decomposes into calcium oxide (\( CaO \)) and carbon dioxide (\( CO_2 \)):
\[ CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g) \]

### Step 2: Calculate the molar mass of \( CaCO_3 \)
The molar mass of \( CaCO_3 \) can be calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 3 oxygen atoms)

Thus, the molar mass of \( CaCO_3 \) is:
\[ 40 + 12 + (3 \times 16) = 100 \text{ g/mol} \]

### Step 3: Determine the number of moles of \( CaCO_3 \) in 1 g
Using the molar mass, we can find the number of moles of \( CaCO_3 \) in 1 g:
\[ \text{Number of moles of } CaCO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{100 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of \( CO_2 \) produced
From the balanced equation, we see that 1 mole of \( CaCO_3 \) produces 1 mole of \( CO_2 \). Therefore, 0.01 moles of \( CaCO_3 \) will produce 0.01 moles of \( CO_2 \).

### Step 5: Calculate the volume of \( CO_2 \) at STP
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of \( CO_2 \) produced can be calculated as:
\[ \text{Volume of } CO_2 = \text{number of moles} \times 22.4 \text{ L/mol} = 0.01 \text{ moles} \times 22.4 \text{ L/mol} = 0.224 \text{ L} \]

### Final Answer
The volume of \( CO_2 \) at STP obtained by heating 1 g of \( CaCO_3 \) is **0.224 liters**.

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