If 603 mol of octane combusts, what volume of carbon dioxide is produced at 28.0 °C and 0.995 atm?

Question

anonymous · Answer

$$2C8H18+25O2\rightarrow16CO2+18H2O$$

matt101 · Answer

The ratio of octane to carbon dioxide is 2:16 = 1:8. Therefore 603 mol of octane produces 4824 mol of carbon dioxide. Now use PV = nRT:

(0.995)V = (4824)(0.08206)(301)
V ~ 120 000 L

anonymous · Answer

It should be noted however @matt101's answer is theoretical, NOT actual.  In the real-world you don't even get close to 100% efficiency.  :-)

I would express as a less than or equal to, just in case.

matt101 · Answer

Absolutely correct @agentx5 - I did a double take when I got 120k L for exactly that reason :)