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To calculate how many grams of quicklime (CaO) can be produced from 1.0 kg of limestone (CaCO3) through heating, we need to use the molar masses and stoichiometry.
First, let's determine the molar masses of the compounds involved:
- The molar mass of limestone (CaCO3) is calculated as follows:
- Molar mass of Ca = 40.08 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + 3*(16.00 g/mol) = 100.09 g/mol
Next, we need to determine the molar ratio between limestone and quicklime:
- The balanced equation for the decomposition of limestone is: CaCO3 → CaO + CO2
- The stoichiometric coefficients indicate that 1 mole of CaCO3 produces 1 mole of CaO.
Now, let's calculate the number of moles of limestone in 1.0 kg:
- Using the molar mass of CaCO3, we can calculate the number of moles:
- Moles of CaCO3 = mass (g) / molar mass (g/mol)
- Moles of CaCO3 = 1000 g / 100.09 g/mol ≈ 9.996 moles
Since the molar ratio between CaCO3 and CaO is 1:1, we can conclude that 1.0 kg of limestone can produce approximately 9.996 moles of CaO.
To calculate the mass of CaO produced, we can use the molar mass of CaO:
- The molar mass of CaO = 40.08 g/mol
Finally, we can calculate the mass of CaO produced:
- Mass of CaO = moles of CaO × molar mass of CaO
- Mass of CaO = 9.996 moles × 40.08 g/mol ≈ 400.57 g
Therefore, approximately 400.57 grams of quicklime (CaO) can be produced from 1.0 kg of limestone (CaCO3) through heating.
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