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.
- Calculate the mass of pure CaCO₃:
\[ \text{Mass of pure CaCO₃} = \text{Total mass} \times \text{Purity} \] \[ \text{Mass of pure CaCO₃} = 20 , \text{g} \times 0.20 = 4 , \text{g} \]
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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:
\[ \text{Moles of CaCO₃} = \frac{\text{Mass of CaCO₃}}{\text{Molar mass of CaCO₃}} = \frac{4 , \text{g}}{100.09 , \text{g/mol}} \approx 0.0399 , \text{mol} \]
- 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:
\[ \text{Moles of CO₂} = 0.0399 , \text{mol} \]
- 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.