[hofmeister4]

"   CH4+2 O2 -> CO2 + 2H2O
A mixture of Methane and Oxygen was prepared from 250ml of methane (measured at 25 Celsius and 745mm Hg) and 300ml of Oxygen (measured at 70 Celsius and 757 mm Hg)
a)   Calculate the Mole Fraction of the gases CH4 and O2 at the initial moment (after mixing both gases and before the start of the reaction) Identify the limiting reactant.
b)   Calculate the maximum volume of CO2 released during the process when it is measured at 1ATM."

I need help with this problem
Specially at the CO2 volume part

[Borek]

You have to show your efforts to get help. This is a forum policy.

[hofmeister4]

Ok, sorry for that, i'm new around here. So, i got CH4+2O2 ->CO2 +2H2O,
PV=nRT 745.0,25=n.62,3.(273+25)

186,25/18565= 0,010 moles CH4

and for O2

757.0,3=n.62,3.(273+70)

n= 0,01 moles O2

So for molar fraction i got nO2=0,01/0,01+0,01 =0,5

and nCh4=0,01/0,01+0,01= 0,5

But i have absolutely no idea on how to do the "B"
i think its missing info on the temperature

[Borek]

Agreed, there is no way to answer the question the way it is worded.

You can find the limiting reactant and calculate the number of moles of CO₂ produced, though. That's what I would do.

[Enthalpy]

Not only is the temperature missing. The reaction would produce much CO instead of CO₂ for lack of oxygen, and would leave some unburnt H₂ too, in proportion not trivial to compute.

Or does anyone imagine that the excess CH₄ would remain untouched in a flame...? Bizarre wording.