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!