When calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂) according to the following chemical reaction:
\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]
To find the total mass of the products, we can follow these steps:
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Calculate the molar mass of CaCO₃:
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol (3 oxygens in CaCO₃)
So,
\[ \text{Molar mass of CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 , \text{g/mol} \]
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Determine the moles of CaCO₃: \[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{30 , \text{g}}{100.09 , \text{g/mol}} \approx 0.2997 , \text{mol} \]
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Use the balanced chemical equation to find the moles of the products: From the reaction, 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂. Therefore, 0.2997 moles of CaCO₃ will produce:
- 0.2997 moles of CaO
- 0.2997 moles of CO₂
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Calculate the mass of the products:
- Molar mass of CaO = 40.08 g/mol (Ca) + 16.00 g/mol (O) = 56.08 g/mol
- Molar mass of CO₂ = 12.01 g/mol (C) + (2 × 16.00 g/mol) = 44.01 g/mol
Now, calculate the mass of each product:
\[ \text{Mass of CaO} = 0.2997 , \text{mol} \times 56.08 , \text{g/mol} \approx 16.80 , \text{g} \]
\[ \text{Mass of CO}_2 = 0.2997 , \text{mol} \times 44.01 , \text{g/mol} \approx 13.18 , \text{g} \]
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Add the masses together to find the total mass of the products: \[ \text{Total mass} = \text{Mass of CaO} + \text{Mass of CO}_2 \approx 16.80 , \text{g} + 13.18 , \text{g} \approx 30.00 , \text{g} \]
Conclusion:
The total mass of the products is approximately 30 grams. According to the law of conservation of mass, the mass of the reactants (30 grams of CaCO₃) equals the mass of the products (CaO and CO₂) upon completion of the reaction.