Question

assuming fully decomposed , the volume of CO2 released at stp of heating on 9.85 gm BACO3 will

Answer 1

BaCO3 - BaO + CO2 ( 197 g) According to the given equation,  1 mole (197 g) of Barium carbonate gives 1 mole (22.4 L) of Carbon di-oxide.  Therefore, 9.85 g of Barium carbonate will give (22.4x9.85) / 197 = 1.12 L of Carbon di-oxide gas at S.T.P