Question

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

Answer 1

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.