In the second experiment, another flask is used for a combustion reaction.
Method
Remove all the air from 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.

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
Original post by cbreezyy_17
In the second experiment, another flask is used for a combustion reaction.
Method
Remove all the air from 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.

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

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.
Original post by PM-G_555
Oxygen is in excess as C8H18 is the limiting reagent, so that adds to the moles of gas left over

Ohh okay makes sense as to why thank you
Original post by TypicalNerd
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
how did you get M1?
Original post by cbreezyy_17
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??
Original post by Pkyoa
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
Original post by TypicalNerd
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

Oooh so it's the amount of oxygen reacting to the 0.001mol of hydrocarbon. Okay that makes more sense thanx
Original post by TypicalNerd
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
Original post by Foyin07
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

Look again at the equation they gave in the question:

C8H18(g) + 12.5O2(g) ——> 8 CO2(g) + 9 H20(l)

For every one mole of the hydrocarbon burned, you use up 12.5 moles of O2, hence the moles of O2 used up = 12.5 x moles of hydrocarbon
Original post by mmm888
how did you get M1?

my boi