When calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂) according to the following chemical reaction:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
In this reaction, the calcium carbonate decomposes into solid calcium oxide and gaseous carbon dioxide. When dealing with chemical reactions, it's crucial to consider the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.
Initially, you have 30 grams of calcium carbonate. The products of the reaction will include the solid calcium oxide and the gaseous carbon dioxide. The total mass of the products will still equal the initial mass of the reactant:
- The mass of the solid CaO will be less than 30 grams (since some mass is lost as CO₂ gas).
- The CO₂ gas will account for the mass lost during the reaction.
However, if you want to find out the exact masses of the products formed, you'd need to calculate the mass of each product using stoichiometry.
To do this, you could first calculate the molar mass of calcium carbonate:
- Ca (Calcium): 40.08 g/mol
- C (Carbon): 12.01 g/mol
- O (Oxygen): 16.00 g/mol × 3 = 48.00 g/mol
Total molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol
Then, you could find the moles in 30 grams of CaCO₃:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{30 , \text{g}}{100.09 , \text{g/mol}} \approx 0.299 , \text{mol} \]
From the reaction, 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂:
- Molar mass of CaO = 40.08 + 16.00 = 56.08 g/mol
- Molar mass of CO₂ = 12.01 + 16.00 × 2 = 44.01 g/mol
Now, calculating the mass of each product formed:
-
Mass of CaO produced: \[ \text{mass of CaO} = \text{moles} \times \text{molar mass} = 0.299 , \text{mol} \times 56.08 , \text{g/mol} \approx 16.77 , \text{g} \]
-
Mass of CO₂ generated: \[ \text{mass of CO}_2 = \text{moles} \times \text{molar mass} = 0.299 , \text{mol} \times 44.01 , \text{g/mol} \approx 13.16 , \text{g} \]
Adding both masses together gives:
\[ \text{Total mass of products} = \text{mass of CaO} + \text{mass of CO}_2 \approx 16.77 , \text{g} + 13.16 , \text{g} = 29.93 , \text{g} \]
So the total mass of the products is approximately 29.93 grams, indicating some mass has been lost as the gas escapes. The close value shows that in a practical setting, the total mass of the products might be expected to be around 30 grams, in line with conservation of mass principles, if taking into account experimental error or rounding.