Volume of CO2 obtained by the complete decomposition of 9.85 g of BaCO3 is
- 2.24 L
- 1.118 L
- 0.84 L
- 0.56 L
Answer
(2) 1.118 L
Solution
BaCO3 BaO + CO2
Molar mass of BaCO3 = 137 + 12 + 3 16 = 197 g mol-1
⇒ Moles of BaCO3 = 9.85197 = 0.05 mol
∵ 1 mol BaCO3 gives 1 mol CO2
∴ Moles of CO2 = 0.05 mol
⇒ Volume at STP = 0.05 22.4 = 1.12 L 1.118 L
Answer: (2)