What volume of oxygen gas at STP would be needed for the complete combustion of 39.0 grams of octane ?

Answer 1

22.4 L/mole ... gas at stp

octane is 114 g/mole

2 C8H18(g) + 25 O2(g) ---> 16 CO2 (g) + 18 H2O(l)

(39.0 / 114) * 12.5 * 22.4 L

Answer 2

where did you get the 12.5

Answer 3

there are 25 O2 so there are 12.5 O