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!

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.

Well, well, well, look who's finally got it! You've cracked the code, my friend! That's absolutely correct! By dividing 312.5 L of pure oxygen by 0.209 (which represents the percentage of oxygen in air), you're essentially canceling out the oxygen, leaving you with the volume of air needed for complete combustion. You've nailed it! Great job, my astute chemistry aficionado! Keep up the good work!

Yes, you are correct! When dividing 312.5 L of O2 by 0.209 (the fraction of oxygen in air), the units of O2 cancel out, leaving you with L. This calculation gives you the total volume of air needed for complete combustion of 25.0 L of octane vapor.

Yes, you got it right! To understand why we divide 312.5 by 0.209, let's break down the problem step by step:

We start with 25.0 L of octane vapor, C8H18, and we want to know how much air is needed for complete combustion.

First, we calculate how much oxygen (O2) would be required if it were pure oxygen. We use the mole ratio between octane and oxygen from the balanced chemical equation for combustion:

C8H18 + 12.5O2 -> 8CO2 + 9H2O

From the equation, we see that 2 moles of octane (C8H18) react with 25 moles of oxygen (O2).

We can convert the given 25.0 L of octane vapor to moles using its molar volume (22.4 L/mol):

25.0 L C8H18 x (1 mol C8H18/22.4 L C8H18) = 1.1167 mol C8H18

Now we can use the mole ratio to find how many moles of oxygen (O2) are needed:

1.1167 mol C8H18 x (25 mol O2/2 mol C8H18) = 13.958 mol O2

So, if we were using pure oxygen, we would need 13.958 moles of O2.

However, air is only 20.9% oxygen by volume. To find out how many liters of air are needed, we need to find the volume of oxygen that corresponds to 13.958 moles if it were pure oxygen.

Let's call the volume of air we need "x" liters. So, 20.9% of "x" liters will be the volume of oxygen:

20.9% of x = (20.9/100) * x = 0.209x

Now we set up an equation to solve for x:

0.209x = 13.958

Dividing both sides by 0.209, we get:

x = 13.958 / 0.209 ≈ 66.85 L

Therefore, you would need approximately 66.85 liters of air for complete combustion of 25.0 liters of octane vapor.

I hope this clarifies your understanding!