2. If air is 20.9% oxygen by volume,

Question

a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?

25L C8H18 x (25 mol O2/2 mol C8H18) = 312.5 L O2 (if it were pure O2).
Then 20.9% of what number = 312.5
0.209x = 312.5

********* I do not understand why you divide 312.5 by 0.209.....so air consists of O2,N2,CO2,....and the question asks how many L of air given that air is 20.9% O2......OH! I THINK I GOT IT NOW....so 312.5 L O2/0.209O2.....the O2 cancel which leaves L... Is that correct????!!!!

Thank you very much!

Answer 1

I don't think it's a matter of units. I just generated then solved the algebraic equation.
312.5 L O2 is required if the O2 used is 100% pure O2. Air is only 20.9% O2 so you know it will take more air. If you want 312.5L pure O2 you want to know how much air to begin with (at 20.9% O2) to obtain the 312.5L needed. That's why I did 20.9% O2 in air x what number = 312.5L O2? Convert that "what number" to a symbol like y and convert 20.9% to a fraction and you get a simple algebraic equation of 0.209*y = 312.5. Solve for y.
y = 312.5/0.209 = about 1495 L air containing 20.9% O2 will give you 312.5 O2 which you need to completely combust the octane.