# Question #00893

Dec 3, 2014

If I understood the question correctly, the volume will be $V = 69.7 L$

Basically, you are dealing with the combustion of benzene, ${C}_{6} {H}_{6}$. Starting from the balanced reaction,

$2 {C}_{6} {H}_{6} + 15 {O}_{2} \to 12 C {O}_{2} + 6 {H}_{2} O$

you can see that the benzene to oxygen ratio is $2 : 15$; that is, for every 2 moles of benzene you need 15 moles of oxygen.

Your initial values are $0.5$ moles for ${C}_{6} {H}_{6}$ and $3.75$ moles for ${O}_{2}$, which correspond to the balanced equation's ratio. This means that neither the benzene, nor the oxygen are limiting reactants.

So, for every $2$ moles of ${C}_{6} {H}_{6}$ you will produce $12$ moles of $C {O}_{2}$ and $6$ moles of ${H}_{2} O$.
GIven that you have $0.5$ moles to start with, you will end up with $3$ moles of $C {O}_{2}$ ( 1:6 ratio ) and $1.5$ moles of ${H}_{2} O$ ( 1:3 ratio).

Now, using the ideal gas law $P V = n R T$ and knowing that you have a total of $3 + 1.5 = 4.5$ moles of gas produced, you get

$V = \frac{n R T}{P} = \frac{4.5 \cdot 0.082 \cdot \left(105 + 273\right)}{2.0} = 69.7 L$