Home barbeques are fueled with propane gas (C_3H_8). What mass of carbon dioxide (in kg) forms upon the complete combustion of 18.9 propane (approximate contents of one 5-gallon tank)?

Assume that density of the liquid propane in the tank is 0.621 g/mL

1 Answer
Aug 11, 2018

Answer:

Well, you have not given us all the necessary info....

Explanation:

We address the combustion equation..

#H_3C-CH_2CH_3(g) + 5O_2(g) rarr 3CO_2(g) +4H_2O(l) + Delta#

And, assuming you are a septic, #"1 US Gallon"-=3.785*L#..

We combust a mass of ethane...

#"5 US" cancel("Gallon")xx3.785*cancelL*cancel"gallon"^-1xx10^3*cancel(mL)*cancel(L^-1)xx0.621*g*cancel(mL^-1)-=11752.4*g#...

...the which represents a molar quantity with respect to ethane of...

#(11752.4*g)/(30.07*g*mol^-1)=390.8*mol#...

And given the combustion, we get THREE equiv of carbon dioxide evolved...

#390.8*molxx3xx44.01*g*mol^-1xx10^3*g*kg^-1=51.6*kg#.

Take that atmosphere...