When solid calcium carbonate (CaCO3) is heated, it decomposes to form solid calcium oxide (CaO) and carbon dioxide gas (CO2). How many liters of carbon dioxide will be produced at STP if 2380 g of calcium carbonate reacts completely? CaCO3 (s) CaO (s) + CO2 (g)

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

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