Question

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

(a) 1.12 L 

(b) 0.84 L

(c) 2.24 L

(d) 4.06 L

Answer 1

Correct option is (a) 1.12 L

BaCO₃​ → BaO + CO_2​

When 197 grams of BaCO₃​ is decomposed it forms 22.4 liters of CO₂​

If 9.85 grams of BaCO₃​ is decomposed it forms X liters of CO_2​.

X = \(\frac{9.85 \times 22.4}{197}\) = 1.12