Question

Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃ (Atomic mass of Ba = 137) will be _______.
(A) 0.84 L
(B) 2.24 L
(C) 4.06 L
(D) 1.12 L

Answer 1

Answer:

Explanation:

Quantity of Baco3 = 9.85g

BaCO3 can be described as -

BaCO3 → BaO+ CO₂2

The atomic Mass of Ba =137

The atomic Mass of C =12

The atomic Mass of O =16

Therefore, the molecular mass of BaCO3

= 137 + 12 + (16 × 3)

= 197

197 gm of BaCO₃ releases carbon dioxide = 22.4 at STP

1 gm of BaCO₃ releases carbon dioxide = 22.4/197  

9.85 gm of BaCO₃ released carbon dioxide

= 22.4/197 × 9.85

= 1.12 litre