To determine the total mass of the products from the chemical reaction involving calcium carbonate (CaCO₃) decomposing into carbon dioxide (CO₂) and calcium oxide (CaO), we first need to write the balanced chemical equation:
\[ \text{CaCO}_3 \rightarrow \text{CO}_2 + \text{CaO} \]
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Determine the molar masses:
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Molar mass of CaCO₃ (calcium carbonate):
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol (3 oxygen atoms)
- Total: \( 40.08 + 12.01 + (16.00 \times 3) = 100.09 \text{ g/mol} \)
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Molar mass of CO₂ (carbon dioxide):
- C: 12.01 g/mol
- O: 16.00 g/mol (2 oxygen atoms)
- Total: \( 12.01 + (16.00 \times 2) = 44.01 \text{ g/mol} \)
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Molar mass of CaO (calcium oxide):
- Ca: 40.08 g/mol
- O: 16.00 g/mol
- Total: \( 40.08 + 16.00 = 56.08 \text{ g/mol} \)
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Calculate the total mass of products: From the reaction:
- 1 mole of CaCO₃ produces 1 mole of CO₂ and 1 mole of CaO.
Therefore, the total mass of the products from the decomposition of 1 mole of CaCO₃ is: \[ \text{Mass of CO}_2 + \text{Mass of CaO} = 44.01 \text{ g/mol} + 56.08 \text{ g/mol} = 100.09 \text{ g/mol} \]
Since the mass of the reactants is equal to the mass of the products (conservation of mass), the total mass of the products from the decomposition of 30 moles of CaCO₃ will be:
\[ 30 \text{ moles} \times 100.09 \text{ g/mol} = 3002.7 \text{ grams} \]
Thus, the total mass of the products from the decomposition of 30 moles of CaCO₃ is 3002.7 grams.