# If 538 mol of octane combusts, what volume of carbon dioxide is produced at 14.0°C and .995 atm?

May 8, 2016

Over one hundred thousands litres of carbon dioxide gas are evolved.

#### Explanation:

Balanced combustion equation:

${C}_{8} {H}_{18} \left(l\right) + \frac{17}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$

So, given that $538 \cdot m o l$ of octane combust, clearly, there are $8 \times 538 \cdot m o l$ $C {O}_{2}$ evolved under complete combustion.

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

$=$

$\frac{8 \times 538 \cdot \cancel{m o l} \times 0.0821 \cdot L \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 287 \cdot \cancel{K}}{0.995 \cdot \cancel{a t m}}$

??*L

What is the volume in ${m}^{3}$?