Question

What is the balanced chemical equation for the decomposition of calcium carbonate and how much CO2 is produced when 150g of CaCO3 decomposes?

Answer 1

The balanced chemical equation for the decomposition of calcium carbonate (CaCO₃) is as follows:

\[
\text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g)
\]

This equation indicates that one mole of calcium carbonate decomposes to produce one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO₂).

To calculate the amount of CO₂ produced from the decomposition of 150 g of CaCO₃, follow these steps:

1. **Calculate the molar mass of CaCO₃:**
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
- Total molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol

2. **Convert grams of CaCO₃ to moles:**
\[
\text{Moles of CaCO}_3 = \frac{150 \, \text{g}}{100.09 \, \text{g/mol}} \approx 1.499 \, \text{moles}
\]

3. **Determine the moles of CO₂ produced:**
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will also be approximately 1.499 moles.

4. **Calculate the mass of CO₂ produced:**
- Molar mass of CO₂:
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol
- Total molar mass of CO₂ = 12.01 + 32.00 = 44.01 g/mol

Using the moles of CO₂ to calculate mass:
\[
\text{Mass of CO}_2 = \text{moles of CO}_2 \times \text{molar mass of CO}_2
\]
\[
\text{Mass of CO}_2 = 1.499 \, \text{moles} \times 44.01 \, \text{g/mol} \approx 65.96 \, \text{g}
\]

Therefore, when 150 g of CaCO₃ decomposes, it produces approximately 66.0 g of CO₂.