The heating of calcium carbonate (CaCO₃) typically leads to a process called thermal decomposition. Upon heating, calcium carbonate decomposes into calcium oxide (CaO) and carbon dioxide (CO₂) gas. The chemical reaction is represented by the following equation:
\[ \text{CaCO}_3 \ (s) \rightarrow \text{CaO} \ (s) + \text{CO}_2 \ (g) \]
To provide additional details about the process:
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Molar Mass Calculation: The molar mass of calcium carbonate (CaCO₃) is about 100.09 g/mol.
- Calcium (Ca) = 40.08 g/mol
- Carbon (C) = 12.01 g/mol
- Oxygen (O) = 16.00 g/mol × 3 = 48.00 g/mol
- Total = 40.08 + 12.01 + 48.00 = 100.09 g/mol
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Moles of Calcium Carbonate: To determine how many moles of calcium carbonate are in 30 grams, use the formula:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{30 , \text{g}}{100.09 , \text{g/mol}} \approx 0.299 , \text{moles} \]
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Products Formed: From the stoichiometry of the reaction, 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂. Therefore, 0.299 moles of CaCO₃ will produce:
- \(0.299 , \text{moles of CaO}\)
- \(0.299 , \text{moles of CO₂}\)
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Mass of Products:
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Mass of CaO produced: \[ \text{Molar mass of CaO} = 40.08 , \text{g/mol} + 16.00 , \text{g/mol} = 56.08 , \text{g/mol} \] \[ \text{Mass of CaO} = 0.299 , \text{moles} \times 56.08 , \text{g/mol} \approx 16.76 , \text{grams} \]
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Mass of CO₂ produced: \[ \text{Molar mass of CO₂} = 12.01 , \text{g/mol} + 16.00 , \text{g/mol} \times 2 = 44.01 , \text{g/mol} \] \[ \text{Mass of CO₂} = 0.299 , \text{moles} \times 44.01 , \text{g/mol} \approx 13.16 , \text{grams} \]
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Conclusion: Upon heating 30 grams of calcium carbonate, approximately 16.76 grams of calcium oxide and 13.16 grams of carbon dioxide are produced, assuming complete conversion and no losses.
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