How many grams of O2 are required to burn 1.20 gal of C8H18

I need help on my chemistry, please! How many grams of O2 are required to burn 1.20 gal of C8H18 if Octane has a density of 0.692 g/mL at 20 degrees C? My last answer was 884 grams. I got this through converting 1.2 gallons to 4542.49 mL. I multiplied that by 0.692 g/mL to get 3143.4 g. And through a (probably very messed up) stoichiometry problem, I got 884 g. Can someone please advise me on how to do this problem? Thank you!!!

Convert 1.2gallons to mL
1.2 gallon [US, liquid] = 4 542.49 cubic centimeter
Calculate mass - density = 0.692g/cm³
Mass = 4,542.49*0.692 = 3,143.4 g. You were correct up to here - good work so far:

Molar mass C8H18 = 114.2 g/mol
moles C8H18 = 3,143.4/114.2 = 27.525 mol

Equation:
2 C8H18 + 25 O2 = 16 CO2 + 18 H2O
2mol C8H18 react with 25mol O2
27.525 mol C8H18 react with 25/2*27.525 = 344.1 mol O2

Molar mass O2 = 32g/mol
344.1 mol = 344.01*32 = 11,010 g O2 required.

You didn't convert it to moles, which is what you need it to be in for stoichiometry.
You also need the balanced stoichiometry to know theoretical ratios.
2C8H18 + 25O2 ---> 16CO2 + 18H2O
Now you know you need 25 moles O2 for 2 moles C818. Since C8H18 has a molar mass of 114.23g, divide that into 3143.4g to get moles.
3143.4g (1mol/114.23g) = 27.52 moles C8H18

Now you can play with the ratio:
2:25 is 1:12.5.
multiply by 27.52 --> 27.52 : 344 moles O2.
Convert this to grams if that's what it's supposed to be in (18g/mol).

Good luck!