In a hypothetical reaction, 18.2 L of carbon dioxide is generated in the decomposition of calcium carbonate. What mass of calcium carbonate is decomposed at STP?

Question

CaCO3 (s) ---> CaO (s) + CO2 (g)

Answer 1

At STP, 1 mole of any ideal gas occupies 22.4 liters. From this, we can determine the moles of CO₂ in 18.2 L.

18.2 L CO₂ x 1 mol / 22.4 L = 0.8125 moles CO₂

Using the stoichiometry of the balanced equation, we can now find the moles of CaCO3 that produced this amount of CO₂:

0.8125 mols CO₂ x 1 mol CaCO₃ / 1 mol CO₂ = 0.8125 mols CaCO₃

Now, we simply use the molar mass of CaCO₃ (100. g / mol) to find the mass:

0.8125 mols CaCO₃ x 100. g / mole = 81.3 g CaCO₃ (3 sig. figs.)