[chanelo]

The problem is:
2C₈H₁₈ + 25O₂  16CO₂ + 18H₂O
If 394 mol octane combusts, what volume of carbon dioxide is produced at 28.0°C and 0.995 atm?

I started by finding out how many moles of carbon dioxide are produced from the combustion of octane by doing 394 mol x (16 mol CO₂ / 2 mol C₈H₁₈) and got 3152 mol CO₂.
Then I did
V = nRT / P
V = (3152)(0.082)(28 + 273) / 0.0995 and I got 7.82 x 10^-4 L. But this is wrong. Can someone tell me where I went wrong?

[EpicWinston]

Well that's odd.

What is the answer suppose to be? I did the problem and I got the same answer. The only flaw I saw was that you divided by 0.0995 instead of 0.995, but we still arrived at the same answer so you probably typed it correctly in your calculator.

[Borek]

I got 7.82 x 10^-4 L.

10^-4? Below 1 mL of gas?

[mamid]

I did it too, and got 7.82 x 10⁴. That's 4, as opposed to -4. It seems you people just read the result wrong.

[EpicWinston]

Yea that's what I meant 10⁴ my bad

[Borek]

To be precise, I got 7.83×10⁴

[EpicWinston]

How did you get that program @Borek?

[Borek]

I wrote it. But you can just buy, see my signature