In the reaction CaCO3(s)⟶CaO(s)+CO2(g) how many liters of carbon dioxide gas, CO2 , measured at STP are produced from the decomposition of 195 g of calcium carbonate, CaCO3 ?

Answer 1

First, write the correctly balanced equation for the reaction taking place:

CaCO₃ ==> CaO + CO₂ ... balanced equation

Next, calculate the moles of CaCO₃ that are reacting (decomposing):

molar mass CaCO₃ = 100.1 g / mol

moles CaCO₃ = 195 g x 1 mol / 100.1 g = 1.95 moles

Convert moles CaCO₃ to moles CO₂ formed (use the balance equation to do this):

1.95 mols CaCO₃ x 1 mol CO₂ / mol CaCO₃ = 1.95 mols CO₂

Finally, convert moles CO2 to liters of CO2:

At STP 1 mole of any ideal gas = 22.4 L (memorize this value)

1.95 mols CO₂ x 22.4 L / mol = 43.7 L (3 sig. figs.)