Answer:
533.12 L
Explanation:
The balanced chemical equation for the decomposition of CaCO₃ is the following:
CaCO₃(s) → CaO(s) + CO₂(g)
According to the equation, 1 mol of CaCO₃ produces 1 mol of gas (CO₂). We convert the moles of CaCO₃ to mass in grams with the molar mass of the compound:
molar mass CaCO₃ = 40 g/mol Ca + 12 g/mol C + (3 x 16 g/mol O)= 100 g/mol
mass of CaCO₃ = 1 mol x 100 g/mol = 100 g CaCO₃
Now, we know that 1 mol of any gas at STP occupies a volume of 22.4 L:
1 mol CO₂ = 22.4 L CO₂
Thus, the stoichiometric ratio is: 22.4 L CO₂/100 g CaCO₃
Finally, we multiply this ratio by the mass of CaCO₃ to calculate how many liters of CO₂ at STP are produced:
2380 g CaCO₃ x 22.4 L CO₂/100 g CaCO₃ = 533.12 L
