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!
Scroll down to your post at time 12:39. I responded to that earlier post.
Yes, you are correct! To find the number of liters of air needed for complete combustion of octane vapor, you need to divide the volume of oxygen required by the fraction of oxygen in air.
In this case, you have calculated that 312.5 L of pure oxygen (if it were pure O2) is needed for complete combustion of 25.0 L of octane vapor. However, the air is only 20.9% oxygen by volume.
To find the volume of air needed, you divide the volume of pure oxygen (312.5 L) by the fraction of oxygen in air (0.209). This is because 0.209 is equal to 20.9% when expressed as a decimal.
So, the calculation would be:
312.5 L O2 / 0.209 O2 = 1495 L air
Therefore, you would need approximately 1495 liters of air for complete combustion of 25.0 liters of octane vapor.
For part (b) of the question, to calculate the volume of each product produced, you would need to use the balanced chemical equation for the combustion reaction of octane. This equation will tell you the stoichiometric ratio between octane and the products formed (such as carbon dioxide and water vapor). By using this ratio, you can calculate the volume of each product produced.