Question

How many moles of CaO are produced when 68.0 g of CaCO3 is heated according to the equation: CaCO3 --> CaO + CO2?

Answer 1

To determine how many moles of CaO are produced from heating 68.0 grams of CaCO3, we need to 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
- Molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol

2. **Calculate the number of moles of CaCO3**:
\[
\text{moles of CaCO3} = \frac{\text{mass of CaCO3}}{\text{molar mass of CaCO3}} = \frac{68.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.678 \, \text{moles}
\]

3. **Use the stoichiometry of the reaction**:
The balanced equation for the decomposition of calcium carbonate is:
\[
\text{CaCO3} \rightarrow \text{CaO} + \text{CO2}
\]
From this equation, we see that 1 mole of CaCO3 produces 1 mole of CaO.

4. **Calculate the moles of CaO produced**:
Since the reaction produces 1 mole of CaO for every mole of CaCO3, the moles of CaO produced is the same as the moles of CaCO3 consumed:
\[
\text{moles of CaO} = \text{moles of CaCO3} \approx 0.678 \, \text{moles}
\]

Therefore, when 68.0 g of CaCO3 is heated, approximately **0.678 moles of CaO** are produced.