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

May 17, 2017

Well we need to find TWO quantities..........and finally get a mass of $11.9 \cdot m g$ with respect to propane.

Explanation:

$\left(i\right)$ $\text{Moles of dioxygen} = \frac{P V}{R T} = \frac{\frac{749.1 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 0.1048 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 328.7 \cdot K}$

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

$= 3.383 \times {10}^{-} 3 \cdot m o l$ WITH RESPECT TO DIOXYGEN.

$\left(i i\right)$ And we follow the stoichiometric reaction to get stoichiometric equivalence,

${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$

And so, $\text{Mass of propane} = \frac{3.383 \times {10}^{-} 3 \cdot m o l}{\frac{25}{2}} \times 44.10 \cdot g \cdot m o {l}^{-} 1$

$= 11.9 \cdot m g$..........

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