Question

assuming full devompostion, the volume of co2 released at stp on heatimg 9.85gm of baco3<br />Baco3 = BaO+Co2

Answer 1

●BaCO3 ==> BaO + CO2

●1 mole of BaCO3 gives 1 mole of CO2

●197.34 gm of BaCO3 gives 44 gm of CO2

●9.85 gm of BaCO3 gives = 9.85×(44/197.34) gm of CO2 = 2.196 gm of CO2

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

●The volume of CO2 obtained at STP by the complete decomposition of 9.85 gm BaSO3 is

= 0.05 × 22.4 liters

= 1.12 liters

= 1.12×1000 ml

= 1120 ml _________[ANSWER]

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_-_-_-_☆$BE \: \: \: \: BRAINLY$☆_-_-_-_

Answer 2

@ Hey Muskan......@
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Have a look for your answer in the above pic dear:---------

For easy to understand,

BaCO3====> BaO + CO2
Molecular mass of BaCO3
----------> 137+12+(3+16)
====>197g
•22.4 L of CO2 is released by 197g of BaCO3
•xL of CO2 is released by 9.85g of BaCO3
So,
X = [22.4 × 9.85]÷197
X = 1.12L = 1120ml------Ans.

Hope it will help you!
Thanks!!
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