Question

The volume (in L) of CO2(g) obtained at STP by completely burning 10 g of 90% pure CaCO3 is approximately (molar volume of CO2 at STP=22.7L) (Assume CO2 as an ideal gas) (CaCO3=100u)

• A. 2.27 L
• B. 3.96 L
• C. 2.044 L
• D. 4.088 L

Hint:

Stoichiometry: moles = mass / MM, adjust for purity. STP molar volume approx 22.4 L, but use given 22.7. Assume complete reaction.

Answer 1

The Correct Option is C

1. The reaction is CaCO$_3 \to$ CaO + CO$_2$ (decomposition by burning).
2. Purity 90%, so effective mass = 0.9 \times 10 = 9 g.
3. Molar mass CaCO$_3$ = 100 g/mol, moles = 9 / 100 = 0.09 mol.
4. 1 mol CaCO$_3$ produces 1 mol CO$_2$, so 0.09 mol CO$_2$.
5. At STP, volume = moles \times 22.7 L/mol = 0.09 \times 22.7 \approx 2.043$ L \approx 2.044 L.
6. Therefore, the correct option is (3) 2.044 L.