Final answer:
To produce 47.0 L of CO2 gas at STP from the decomposition of calcium carbonate, 209.85 grams of calcium carbonate are required.
Step-by-step explanation:
Calcium Carbonate Decomposition
When calcium carbonate (CaCO₃) decomposes upon heating, it yields calcium oxide (CaO) and carbon dioxide gas (CO₂). To determine how many grams of calcium carbonate are needed to produce 47.0 L of carbon dioxide at Standard Temperature and Pressure (STP), we can use the ideal gas law and the stoichiometry of the balanced chemical reaction:
CaCO₃ (s) → CaO(s) + CO₂(g)
At STP (0°C and 1 atm), 1 mole of gas occupies 22.4 L. Therefore, to find the number of moles of CO₂ that correspond to 47.0 L, we divide the volume by 22.4 L/mol.
The calculation is as follows:
47.0 L CO₂ × (1 mol/22.4 L) = 2.0982 moles of CO₂
Since the molar ratio of CaCO₃ to CO₂ is 1:1, we need 2.0982 moles of CaCO₃. To find the mass of CaCO₃ needed, we multiply the number of moles by the molar mass of CaCO₃ (100.09 g/mol):
2.0982 moles × 100.09 g/mol = 209.85 grams of CaCO₃
Therefore, you would need 209.85 grams of calcium carbonate to produce 47.0 L of carbon dioxide gas.