Question

what volume of co2 is obtained when 20 g of 20% pure caco3 is completely heated

Answer 1

To determine the volume of CO₂ obtained when a certain amount of CaCO₃ (calcium carbonate) is heated, we first need to calculate how much pure CaCO₃ is present in the 20 g of a 20% solution.

1. **Calculate the mass of pure CaCO₃**:

\[
\text{Mass of pure CaCO₃} = \text{Total mass} \times \text{Purity}
\]
\[
\text{Mass of pure CaCO₃} = 20 \, \text{g} \times 0.20 = 4 \, \text{g}
\]

2. **Determine the moles of CaCO₃**:
The molar mass of CaCO₃ is approximately:
- Ca (Calcium) = 40.08 g/mol
- C (Carbon) = 12.01 g/mol
- O (Oxygen) = 16.00 g/mol × 3 = 48.00 g/mol

\[
\text{Molar mass of CaCO₃} = 40.08 + 12.01 + 48.00 \approx 100.09 \, \text{g/mol}
\]

Now, we calculate the moles of CaCO₃:

\[
\text{Moles of CaCO₃} = \frac{\text{Mass of CaCO₃}}{\text{Molar mass of CaCO₃}} = \frac{4 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.0399 \, \text{mol}
\]

3. **Determine the stoichiometry of the reaction**:
The decomposition of calcium carbonate (CaCO₃) upon heating is represented by the following reaction:

\[
\text{CaCO₃} \rightarrow \text{CaO} + \text{CO₂}
\]

From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂.

Therefore, the moles of CO₂ produced will be the same as the moles of CaCO₃ decomposed:

\[
\text{Moles of CO₂} = 0.0399 \, \text{mol}
\]

4. **Calculate the volume of CO₂ at standard temperature and pressure (STP)**:
At STP (Standard Temperature and Pressure), 1 mole of gas occupies approximately 22.4 liters.

\[
\text{Volume of CO₂} = \text{Moles of CO₂} \times 22.4 \, \text{L/mol}
\]
\[
\text{Volume of CO₂} = 0.0399 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.893 \, \text{L}
\]

Thus, the volume of CO₂ obtained when 20 g of 20% pure CaCO₃ is completely heated is approximately **0.893 liters** or **893 mL**.