2. If air is 20.9% oxygen by volume,
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!
1 answer
Scroll down to your post at time 12:39. I responded to that earlier post.
3 answers