stoichiometry of gases

Question

4. How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP?

Answer 1

the underlying chemical equation that you're working with here is

CaCO3​(s) → CaO(s) + CO2​(g)

meaning that the mole ratio between elements is 1:1

at STP, one mole of an ideal gas occupies 22.4 L, meaning that 5.00 L is

5.00 L / 22.4 L/mol = 0.223 mol CO2

which is also 0.223 mol CaCO3 since the mole ratio is 1:1

finally then you can multiply by the molar mass to get the end value

0.223 mol CaCO3 x 100.9 g/mol = 22.34 g

so you must decompose 22.34 g of calcium carbonate

Answer 2

Use the unit factor method along with the molar volume @STP of any ideal gas being 22.4 L.

The balanced equation for this decomposition is CaCO3 yields CaO+CO2

?gCaCO3=5.00LCO2x(1MolCO2/22.4LCO2)(1MolCaCO3/1MolCO2)(100gCaCO3/1MolCaCO3)=22.3gCaCO3