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

Correct option: (D) 1.12 L

BaCO₃ → BaO + CO₂↑

Molecular weight of BaCO₃

= 137 + 12 + (3 x 16)

= 197

22.4 L of CO₂ is released by 197 g of BaCO₃

∴ x L of CO₂ is released by 9.85 g of BaCO₃

x = \(\cfrac{22.4\, \times\, 9.85}{197}\) = 1.12 L