Sol 2.
As Combustion Reaction of propane , C3H8 :
C3H8 + 5O2 -----> 3CO2 + 4H2O
As Mass of C3H8 = 20 lb
= 20 × 453.592 g
= 9071.84 g
Molar Mass of C3H8 = 44.1 g/mol
So , Moles of C3H8 = 9071.84 g / 44.1 g/mol
= 205.7106 mol
At STP , 1 mol = 22.414 L
Therefore , Liters of propane gas , C3H8 released at STP
= 205.7106 × 22.414 L
= 4610.80 L
Now , From reaction , Moles of CO2 = Moles of C3H8 × 3
= 205.7106 mol × 3
= 617.1318 mol
Therefore , Liters of CO2 gas produced at STP
= 617.1318 × 22.414 L
= 13832.39 L
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