To determine how many moles of CaO are produced from heating 68.0 grams of CaCO3, we need to 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
- Molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol
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Calculate the number of moles of CaCO3: \[ \text{moles of CaCO3} = \frac{\text{mass of CaCO3}}{\text{molar mass of CaCO3}} = \frac{68.0 , \text{g}}{100.09 , \text{g/mol}} \approx 0.678 , \text{moles} \]
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Use the stoichiometry of the reaction: The balanced equation for the decomposition of calcium carbonate is: \[ \text{CaCO3} \rightarrow \text{CaO} + \text{CO2} \] From this equation, we see that 1 mole of CaCO3 produces 1 mole of CaO.
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Calculate the moles of CaO produced: Since the reaction produces 1 mole of CaO for every mole of CaCO3, the moles of CaO produced is the same as the moles of CaCO3 consumed: \[ \text{moles of CaO} = \text{moles of CaCO3} \approx 0.678 , \text{moles} \]
Therefore, when 68.0 g of CaCO3 is heated, approximately 0.678 moles of CaO are produced.