Question

What is the Volume of co2 obtained at STP by the complete decomposition of 9.85 g BaC03

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

Hope it helps