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Volume of CO2 obtained at STP by the complete decomposition of 9.85 g BaCO3 is?
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Volume of CO2 obtained at STP by the complete decomposition of 9.85 g ...
BaCO3⇒ BaO + CO2

1 mole of BaCO3 gives 1 mole of CO2

197.34 g of BaCO3 gives 44 g of CO2 

9.85 g of BaCO3 gives how many grams of CO2? 

=9.85x44/197.34

= 2.196 g of CO2

No.of moles of CO2=2.196/44=0.05 moles.

The volume of co2 obtained at STP by the complete decomposition of 9.85 g BaC03 is

=0.05x22.4 lts
=1.12 litres 
=1.12x1000 ml
=1120 ml
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Volume of CO2 obtained at STP by the complete decomposition of 9.85 g BaCO3 is?
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