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

2C8H18 + 25O2 > 16CO2 + 18H2O 

2 moles of octane gives 16 moles of CO2 

so 1 mole of octane will give = 16/2 = 8 moles of CO2 

so 603 moles of octane will give 8 X 603 = 4824 moles of CO2 

now use PV = nRT 

P = 0.995 atm 
V = ? L 
n = 4824
R = 0.0821 L atm/K/mole 
T = 28 + 273 = 301 K 

0.995 X V = 4824 X 0.0821 X 301 

V = 4824 X 0.0821 X 301/ 0.995 = 119810.22 L

257 grams h2o i think