Question

volume of CO2 obtained at STP by the complete decomposition of 9.8 g BaCO3 is

Answer 1

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BaCO3=BaO+CO2

no. of moles of BaCO3= mass given/molecular mass

=9.85/197=1/20 moles


As one mole of BaCO3 gives one mole of CO2

thus 1/20 moles of BaCO3 gives 1/20 moles of CO2 mole of CO2

=volume of CO2/volume at STP(22.7lit..)


the equation is 1/20=x/22.7


x=22.7/20=1.135 lit.