Assuming fully decomposed, the volume of  CO2released at STP on heating 9.85 g of BaCO3 (Atomic mass of Ba = 137) will be

BaCO3  →  BaO  +  CO2 ↑   1 mole                             1 moleMolecular  weight of BaCO3 =137 + 12  +  3  ×  16=197∵  197  g of BaCO3 produces  22.4  L ofCO2 at S.T.P.∴  9.85 g  of BaCO3 produces 22.4197 ×  9.85  CO2  =  1.12  L CO2  at S.T.P.