[AWK]

1 mole of mixture contained x mole of CO2 and 1-x mole of CO showing mean molecular mass of 31.32

[Rutherford]

Okay, then x=0.2075 n(O2)=0.2075/2 n(CO)=0.2075+0.2075
CO/O₂=4  

[AWK]

Moles CO before reaction = 1 (CO and CO2 after reaction)

[Rutherford]

1/(0.2075/2)=9.64 wrong answer.
I am starting to think that this can't be solved after all.

[Borek]

It can be solved, 11th post by AWK precisely explains how to do it.

[Rutherford]

I really don't understand. Could AWK or you explain it more clearly. The answer should be 10.3.

[AWK]

First of all - mean molecular mass of air is 28.8 (for 20 % vol. O₂ and 80 % vol. N₂. This gives 0.194 moles of CO₂. And this is your problem. The last your solution is correct, when you use the correct number 28.8 instead of 29 and 31.104 instead of 31.32.

[Rutherford]

I begun to understand now, thanks very much. Just one last question:

Moles CO before reaction = 1 (CO and CO2 after reaction)

why did you use 1 mol to be the n of CO, if you used some other number like 2,3,4,5... would you get the same right?
I know that you said "1 mol of mixture"

[AWK]

Mean molecular mass of gases is calculate for 1 mole of gaseous molecules

[Rutherford]

I never done a problem that way so it is a harder for me to understand:

1 mole of mixture contained x mole of CO2 and 1-x mole of CO showing mean molecular mass of 31.32

I didn't understand where from did you get the 1 mol for this mixture, I though that you could use for example another number like 2:
2 mole of mixture-->x mole of CO2 and 2-x mole of CO
Why 1 if hte mixture doesn't have a molar mass.

[Borek]

It doesn't have to be 1 mole. It is easier to do calculations for 1 mole, but you don't have even to assume number of moles, you can do calculations using molar fractions. However, molar fractions add to 1, so math would be identical.

[Rutherford]

That I wanted to know. So, maybe the easiest way is to assume the number of moles and then to do the way AWK explained.
I thank you two for having patience with me and for helping me to solve and understand this .