What would be the volume of this `CO_(2)` at STP?
Equation : `CaCO_(3) to CaO + CO_(2) (g) (R.A.M. : Ca = 40, C = 12, O = 16)`

Explanation

To find the volume of \( CO_2 \) at STP (Standard Temperature and Pressure) from the decomposition of \( CaCO_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The decomposition of calcium carbonate (\( CaCO_3 \)) can be represented as:
\[ CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g) \]

2. **Identify Moles from the Balanced Equation:**
From the balanced equation, we see that 1 mole of \( CaCO_3 \) produces 1 mole of \( CO_2 \). Therefore:
\[ 1 \text{ mole of } CaCO_3 \rightarrow 1 \text{ mole of } CO_2 \]

3. **Use the Ideal Gas Law for Volume Calculation:**
At STP, 1 mole of any gas occupies a volume of 22.4 liters. Thus, the volume of \( CO_2 \) produced can be calculated as:
\[ \text{Volume of } CO_2 = \text{moles of } CO_2 \times 22.4 \text{ L/mole} \]

4. **Substituting the Values:**
Since we have 1 mole of \( CO_2 \):
\[ \text{Volume of } CO_2 = 1 \text{ mole} \times 22.4 \text{ L/mole} = 22.4 \text{ L} \]

5. **Final Answer:**
Therefore, the volume of \( CO_2 \) at STP is:
\[ \text{Volume of } CO_2 = 22.4 \text{ liters} \]

### Summary:
The volume of \( CO_2 \) produced from the decomposition of \( CaCO_3 \) at STP is **22.4 liters**.