99 Percentile Qs Bank for JEE MainChemistrySome Basic Concepts of Chemistry
When 10   g of 90 % pure limestone is heated, the approximate volume (in L ) of CO 2 liberated at STP is:
Options
- A4 . 4
- B2 . 0
- C4 . 0
- D22 . 4
Correct answer
B. 2 . 0
Step-by-step solution
The decomposition reaction of CaCO 3 is, CaCO 3   ( s ) ⟶ ∆ CaO ( s ) + CO 2   ( g ) At STP , heating 1 mole of CaCO 3 (i.e., 100   g ) liberates 1 mole or 22 . 4   L of CO 2 . 10   g of 90 % pure limestone = 10 × 90 100 = 9   g of pure CaCO 3 . Since, 100   g → 22 . 4   L Therefore, 9   g of CaCO 3 will liberate = 9 × 22 . 4 100 L = 2 . 0   L   CO 2