Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃(Atomic mass Ba=137) will be

Options

(a) 2.24 L
(b) 4.96 L
(c) 1.12L
(d) 0.84L

Correct Answer:

1.12L

Explanation:

BaCO₃ → BaO+ CO₂
Atomic Mass of Ba=137
Atomic Mass of C=12
Atomic Mass of O=16
Molecular Mass of BaCO₃ = 137 + 12 + (16*3) = 197
197 gm of BaCO₃ released carbondioxide = 22.4 litre at STP
1 gm of BaCO₃ released carbondioxide = 22.4/197 litre
9.85 gm of BaCO₃ released carbondioxide = 22.4/197 x 9.85 = 1.12 litre