The other two problems that I think you submitted ( above this one ) are simple stoichiometry and you should be able to solve them yourself. This one is different.
2CH3OH + 3O2 ---> 2CO2 + 4H2O
(place molar coefficients directly before species)
We will now use the ideal gas equation to get moles of methyl alcohol vapor and oxygen. PV = nRT
(1 atm)(1.00 L CH3OH) = n(0.08206 L*atm/mol*K)(273.15 K)
n = 1.00/22.415
= 0.0446 moles CH3OH
(1 atm)(20.0 L O2) = n(0.08206 L*atm/mol*K)(273.15 K)
n = 20/22.415
= 0.8923 moles O2
------ Now, find limiting reactant. I suspect CH3OH limits.
0.0446 moles CH3OH (3 mol O2/2 mol CH3OH)
= 0.0669 moles
------You do not have that many moles CH3OH, so CH3OH limits the reaction and will drive the reaction. ( Try O2 and see that it is in excess )
------Now, drive the reaction to get moles water.
0.0446 moles CH3OH (4 mol H2O/2 CH3OH)
= 0.0892 moles water
------ Once more with the ideal gas equation.
(1 atm)( L) = (0.0892 moles H2O)(0.08206 L*atm/mol*K)(273.15 K)
L = 1.999/1
= 2.00 Liters water produced ( final answer )
======( the limiting reactant and the excessive reactant is noted above )