How much octane could a #0.500*L# volume of air completely combust?

1 Answer
May 17, 2017

Answer:

Well we need to find TWO quantities..........and finally get a mass of #11.9*mg# with respect to propane.

Explanation:

#(i)# #"Moles of dioxygen"=(PV)/(RT)=((749.1*mm*Hg)/(760*mm*Hg*atm^-1)xx0.1048*L)/(0.0821*(L*atm)/(K*mol)xx328.7*K)#

(Note that I did adjust the volume of dioxygen to account for its percentage volume in air.)

#=3.383xx10^-3*mol# WITH RESPECT TO DIOXYGEN.

#(ii)# And we follow the stoichiometric reaction to get stoichiometric equivalence,

#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

And so, #"Mass of propane"=(3.383xx10^-3*mol)/(25/2)xx44.10*g*mol^-1#

#=11.9*mg#..........

Note that we assumed complete combustion, which is a not altogether reasonable assumption.