If 4.10 × 102 mol of octane combusts, what volume of carbon dioxide is produced at 22.0 °C and 0.995 atm?

1 answer

2C8H18 + 25O2 ==>16CO2 + 18H2O

Convert 410 mols C8H18 to mols CO2.
410 x (16 mols CO2/2 mols C8H18) = n
Substitute n into PV = nRT at the conditions listed and solve for volume in liters.