Need help with amount of sub

In the second experiment, another flask is used for a combustion reaction.
Method
Remove all the air from the flask.
Add 0.0010 mol of 2,2,4-trimethylpentane (CaHi) to the flask.
Add 0.0200 mol of oxygen to the flask.
Spark the mixture to ensure complete combustion.
Cool the mixture to the original temperature.
The equation is
C8H18(g) + 12.5O2(g) ——> 8 CO2(g) + 9 H20(l)
Calculate the amount, in moles, of gas in the flask after the reaction.
The answer is 0.0155 mol
M1 amount of CO2 formed in flask = 0.008 mol
Allow ECF from M1 to M2
M2 amount of gas in flask
= 0.0075 (O2) + 0.0080 (M1) = 0.0155 mol
I don’t understand why we multiply 12.5 and 0.001. I watched a YouTube video explaining the question and they mentioned it was to get the same ratio but in an exam I would have done 0.02/12.5 to get the ratio the same. Also by multiplying the way they have done to get the moles won’t this effect the moles of co2

Let’s start with the obvious. Each mole of C8H18 burned makes 8 moles of CO2, so there must be 8 x 0.0010 mol = 0.0080 mol of CO2 gas.
But the equation tells you that for every mole of C8H18 you burn, 12.5 moles of O2 are used up as well. We can deduce that C8H18 is a limiting reactant as 0.0200 mol is more than 12.5 times 0.0010 mol, which means there will be leftover O2 but no leftover C8H18. We can deduce the moles of O2 that were used up must be
12.5 x 0.0010 mol = 0.0125 mol
So the moles of O2 left are
0.0200 mol - 0.0125 mol = 0.0075 mol
Add the numbers of moles of gases together and you have the required 0.0155 mol.

thank you for ur reply !!
I got the first mark but the way I saw the rest of the question was the reaction between c8H18 and 02 aren’t in a 1:1 ratio so I divided 0.02 /12 =1.6667 x 10-3 and then did 0.02-1.6667 x 10-3 and then added it on to 0.08.
If you can can you explain why my method Dosnt work

Dividing by 12 itself is one of the problems with your approach. Look again at the equation - the mole ratio of C8H18 to O2 (in the form 1 : n) is 1 : 12.5, so you’d divide the moles of O2 by 12.5 if you were to do this at all.
The second problem is that this is more of a test to see if you have a limiting reactant or not. The division by 12.5 simply is working out how many moles of C8H18 it would take to react with all the O2. This test shows that you would need 0.0016 moles of C8H18 to react completely with the 0.0200 moles of O2, which exceeds the amount of C8H18 you have, meaning the oxygen is in excess and the C8H18 is the limiting reagent. What you have done is calculated a theoretical amount of C8H18 and subtracted that from the original number of moles of O2, which is a physically meaningless quantity.