Assertion (A) : 100g of ${\text{CaCO}}_{\text{3}}$ gives 22.4 lit of ${\text{CO}}_{\text{2}}$ at STP

Reason (R) : 100g of ${\text{CaCO}}_{\text{3}}$ gives 44g of CaO       

Balanced Chemical Equation

The thermal decomposition of calcium carbonate is:

CaCO3(s) → CaO(s) + CO2(g)

It is a 1:1:1 mole ratio. One mole of CaCO₃ gives one mole of CaO and one mole of CO₂.

Key Stoichiometry and Molar Mass

At STP, one mole of any ideal gas occupies 22.4 L.

Check the Assertion (A)

1. Moles of CaCO₃ in 100 g = 100 g ÷ 100.09 g/mol ≈ 0.999 mol ≈ 1 mol.
2. Moles of CO₂ formed = 1:1 ⇒ ≈ 1 mol.
3. Volume of CO₂ at STP = 1 mol × 22.4 L/mol = 22.4 L.

Verdict: The assertion is true to the intended significant figures used in school problems. Using exact molar masses gives 22.38 L, which rounds to 22.4 L. So A is correct.

Check the Reason (R)

1. Moles of CaO produced from 100 g CaCO₃ = ≈ 1 mol (1:1 ratio).
2. Mass of CaO = moles × molar mass = 1 × 56.08 g/mol = ≈ 56 g.

Verdict: The reason is false. 100 g CaCO₃ does not give 44 g CaO. It gives about 56 g CaO. The value 44 g matches the molar mass of CO₂, not CaO.