Question

Volume of CO2 obtained by the complete decomposition of 9.85 gm. BaCO3 is :

Answer 1

This question is from stoichiometry the reaction that takes place is BaCO3=BaO+CO2no. of moles of BaCO3= mass given/molecular mass=9.85/197=1/20 moles As one mole of BaCO3 gives one mole of CO2thus 1/20 moles of BaCO3 gives 1/20 moles of CO2 mole of CO2=volume of CO2/volume at STP(22.7l)the equation is 1/20=x/22.7 x=22.7/20=1.135l