A level chemistry help please

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 get M1 but the second part I don’t understand where the 0.0075 comes from Ik it says oxygen in brackets but O2 value is 0.0200mol and why would there be oxygen after the reaction anyway cuz in the equation only carbon dioxide and water is left. Someone please explain thank youu

Oxygen is in excess as C8H18 is the limiting reagent, so that adds to the moles of gas left over

The oxygen is in excess. That means some of it will be left over after the reaction.

You initially have 0.02 mol of O2.
You also initially have 0.001 mol of C8H18.

Using the equation and the ratio of C8H18:O2, how many moles of oxygen are needed? Subtract this number from the initial number of moles of O2.

Thank you sm for explaining it I get it now 0.001*12.5 = 0.0125 moles. 0.02-0.0125 = 0.0075 moles

Why would you do 0.001*12.5 and not 0.02*12.5??

The oxygen is in excess.

Because there are 0.001 moles of hydrocarbon and in the equation, for every 1 mol of hydrocarbon that reacts, 12.5 mol of oxygen react, so the mol of oxygen that reacts = 12.5 x 0.001 mol = 0.0125 mol

The oxygen is in excess.

Because there are 0.001 moles of hydrocarbon and in the equation, for every 1 mol of hydrocarbon that reacts, 12.5 mol of oxygen react, so the mol of oxygen that reacts = 12.5 x 0.001 mol = 0.0125 mol

Hi! why did you use 12.5
i thought it would be 6 since it’s 12 and a half and when i write that in my calculator it says 6

how did you get M1?

can someone go through the entire question please I really don't get it! thank you,

We are given the following information:
-The equation is C8H18 (g) + 12.5O2 (g) —> 8CO2 (g) + 9H2O (l)
-We initially have 0.0010 mol of C8H18 and 0.0200 mol of O2.
From the equation, we can see that for every mol of C8H18 that reacts, we use up 12.5 mol of oxygen.
From this, we can find out how much oxygen needs to react in order for all of the C8H18 to be used up - 12.5 x 0.0010 = 0.0125 mol. This would of course leave us with 0 mol of C8H18 and so we don’t need to add any moles of this to the total at the end.
This is not all of the oxygen, as we had 0.0200 mol to begin with. This means that 0.0075 mol of the oxygen must be left, since 0.0200 mol - 0.0125 mol = 0.0075 mol. As oxygen is a gas as stated in the equation, it does need to be added to the total at the end.
But if we look again at the equation, we see that one of the products (carbon dioxide) is a gas and so we have to also find the moles of this product produced.
We can see from the equation that for every C8H18 that reacts, you make 8 CO2 molecules, so 8 x 0.0010 = 0.0080 mol of CO2.
So the total amount of gas made is 0.0080 mol + 0.0075 mol = 0.0155 mol

The oxygen is in excess. That means some of it will be left over after the reaction.
You initially have 0.02 mol of O2.
You also initially have 0.001 mol of C8H18.
Using the equation and the ratio of C8H18:O2, how many moles of oxygen are needed? Subtract this number from the initial number of moles of O2.