# Question e3578

Aug 17, 2015

You'd need 1300 L.

#### Explanation:

Start by having a look at the balance d chemica lequation for this combustion reaction

$\textcolor{red}{2} {C}_{8} {H}_{18 \left(g\right)} + \textcolor{b l u e}{25} {O}_{2 \left(g\right)} \to 16 C {O}_{2 \left(g\right)} + 18 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $\textcolor{red}{2} : \textcolor{b l u e}{25}$ mole ratio between octane and oxygen gas. This means that for every mole of octane that you want to burn, you need $\frac{25}{2}$ moles of oxygen.

You know that you want to burn 4 moles of octane, which means that you're going to need

4color(blue)(cancel(color(black)("moles octane"))) * (color(blue)(25)" moles O"""_2)/(color(red)(2)color(blue)(cancel(color(black)("moles octane")))) = "50 moles O"""_2#

Now all you have to do is use the ideal gas law equation to determine what volume of oxygen would contain this many moles at $\text{0.953 atm}$ and ${35}^{\circ} \text{C}$.

$P V = n R T \implies V = \frac{n R T}{P}$

${V}_{{O}_{2}} = \left(50 \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.082(color(blue)(cancel(color(black)("atm"))) * "L")/(color(blue)(cancel(color(black)("mole"))) * color(blue)(cancel(color(black)("K")))) * (273.15 + 35)color(blue)(cancel(color(black)("K"))))/(0.953color(blue)(cancel(color(black)("atm}}}}\right)$

${V}_{{O}_{2}} = {\text{1325.7 L O}}_{2}$

Now, I'll leave the answer rounded to two sig figs, even though you only gave one sig fig for the moles of octane

${V}_{{O}_{2}} = \textcolor{g r e e n}{\text{1300 L}}$