A scientist heats 30 grams of calcium carbonate (

1 answer

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:

  1. 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
  2. Moles of Calcium Carbonate: To determine how many moles of calcium carbonate are in 30 grams, use the formula:

    \[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{30 , \text{g}}{100.09 , \text{g/mol}} \approx 0.299 , \text{moles} \]

  3. 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:

    • \(0.299 , \text{moles of CaO}\)
    • \(0.299 , \text{moles of CO₂}\)
  4. Mass of Products:

    • Mass of CaO produced: \[ \text{Molar mass of CaO} = 40.08 , \text{g/mol} + 16.00 , \text{g/mol} = 56.08 , \text{g/mol} \] \[ \text{Mass of CaO} = 0.299 , \text{moles} \times 56.08 , \text{g/mol} \approx 16.76 , \text{grams} \]

    • Mass of CO₂ produced: \[ \text{Molar mass of CO₂} = 12.01 , \text{g/mol} + 16.00 , \text{g/mol} \times 2 = 44.01 , \text{g/mol} \] \[ \text{Mass of CO₂} = 0.299 , \text{moles} \times 44.01 , \text{g/mol} \approx 13.16 , \text{grams} \]

  5. 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.

If you have specific questions or need further clarifications, feel free to ask!