Chemistry problem?

Question

This is the problem I am having trouble with, if you could show me step by step how to do this I would be so thankful.

When 500.0 g of octane, C8H18, is completely burned, what volume of CO2(g) is produced at standard atmospheric pressure and a temperature of 18 degrees C?

The unbalanced equation for the reaction is:

C8H18(l) + O2(g) ---- CO2(g) + H2O(g)

Answer 1

1. Convert grams to moles octane.
2. Balance the equation.
3. Convert moles of CO2 to liters, using the ideal gas law:
PV=nRT, where:

P= 1 atm,
V=unknown,
n= moles,
R = constant (look it up),
T = (18 + 273.15 degrees K).

Solve for V.

Done!

Answer 2

Wow, I didn't know just giving the answer was considered step by step.

Balance the equation:
C8H18 + 25/2 O2 ---> 8 CO2 + 9 H20 (or 2, 25, 16, 18, if you prefer, but that's not important to answering the question)

So, for every mole of octane, you get 8 moles CO2. 500.0 g/114 g/mol of octane = 4.386 moles. So, you will get 35.09 moles.

Plug that into the Ideal Gas Law equation, and solve for volume.

V = nRT/P = 35.09 moles * 0.08206 * 291 K/1 atm = 837.9 L.

Answer 3

Answer
838.28 L