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:
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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
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Convert grams of CaCO₃ to moles: \[ \text{Moles of CaCO}_3 = \frac{150 , \text{g}}{100.09 , \text{g/mol}} \approx 1.499 , \text{moles} \]
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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.
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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: \[ \text{Mass of CO}_2 = \text{moles of CO}_2 \times \text{molar mass of CO}_2 \] \[ \text{Mass of CO}_2 = 1.499 , \text{moles} \times 44.01 , \text{g/mol} \approx 65.96 , \text{g} \]
- Molar mass of CO₂:
Therefore, when 150 g of CaCO₃ decomposes, it produces approximately 66.0 g of CO₂.