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

1 Answer
anor277
May 17, 2017

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=PVRT=749.1⋅mm⋅Hg760⋅mm⋅Hg⋅atm−1×0.1048⋅L0.0821⋅L⋅atmK⋅mol×328.7⋅K

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

=3.383×10−3⋅mol WITH RESPECT TO DIOXYGEN.

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

C8H18(l)+252O2(g)→8CO2(g)+9H2O(l)

And so, Mass of propane=3.383×10−3⋅mol252×44.10⋅g⋅mol−1

=11.9⋅mg..........

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

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