Question

assuming fully decomposed the volume of Co2 relesed at stp on heating 9.85g of BaCo3 (atomic mass, Ba=137)​

Answer 1

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?

=9.85×44/197.34

= 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

=0.05×22.4 lts

=1.12 litres

=1.12×1000 ml

=1120 ml

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