volume of CO2 obtained at STP by complete decomposition of 9.85 Gram barium carbonate is
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Baco3= Ba0+CO2
At first we have to find mass of CO2,
so,. 9.85/275=w/44
w =9.85×44/257
w =433.4/257
w =1.68
now,
mole of CO2= mass/ molar mass
= 1.68/44
= 0.038
now, vol= mol × 22.4
=0.038×22.4
=0.85 litre
At first we have to find mass of CO2,
so,. 9.85/275=w/44
w =9.85×44/257
w =433.4/257
w =1.68
now,
mole of CO2= mass/ molar mass
= 1.68/44
= 0.038
now, vol= mol × 22.4
=0.038×22.4
=0.85 litre
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