# If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

Jul 4, 2017

Well we will treat gasoline as octanes......${C}_{8} {H}_{18}$, and we get a volume of

#### Explanation:

We need (i) a stoichiometric equation........

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

And so we need $\frac{25}{2}$ equiv dioxygen gas.......

And thus if $4 \cdot m o l$ of octane are combusted we need $\frac{25}{2} \times 4 = 50 \cdot m o l$ $\text{dioxygen gas.}$

And (ii) given $V = \frac{n R T}{P} = \frac{50 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 308 \cdot K}{0.953 \cdot a t m}$

Approx....$1300 \cdot L$ or $1.3 \cdot {m}^{3}$ $\text{dioxygen gas}$.