Assuming fully decomposed the volume of co2 released at stp on heating 9.85g baco3
Answers
Answer :
1120 ml
Explanation :
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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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