[Shing]

A 20.0 L container was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture producing water. What is the pressure in the tank at 25degrees C?

What I have done so far:

2H₂ + O₂ --> 2H₂O

P_hydrogen = 202.6KPA
P_Oxygen=303.6 KPA
T= 298 K

Total moles present in the mixture: PV=nRT
                                                n = PV/RT = 506 KPA (the combined pressure) * (20.0)/(8.31)*(298 )
= 4.09 mol


moles of H₂ present = 2/18 *4.09 = 0.45 mol
moles of O₂ present = 16/18 *4.09 = 3.59 mol
H is the limiting reactant.

Moles of O₂ used = 2/1 : 0.45/x
x= 0.225 mols of O₂ are used up

so, moles of 0₂ left are : 3.365 mol (original mol - mol used)

Pressure of O₂ unused = PV=nRT
                                                 P= nRT/V
                                                 = 3.365 * 8.31* 298/ 20.0 l
                                                 = 416.6 Kpa

moles of water formed = 0.45 mol
so how do i find the vapour pressure since H₂O (g) is not ideal and i cant use PV=nRT...
was anything I did right?
I think all is wrong
Helpp!!!

[Borek]

moles of H₂ present = 2/18 *4.09 = 0.45 mol
moles of O₂ present = 16/18 *4.09 = 3.59 mol

This is wrong - use just PV=nRT to caculate separately amount of hydrogen and oxygen. No idea what and why you tried here.

Idea of doing the calculation by finding limiting reagent is OK.

so how do i find the vapour pressure since H₂O (g) is not ideal and i cant use PV=nRT

First of all -  water vapor is close enough to ideal at this temperature.

Then at 25 deg C H₂O is not gas.

Then I have no idea what you are expected to do - could be you are to ignore presence of water vapor. In general this is wrong and you should check saturated vapor pressure in tables, but this thing is sometimes ignored on the HS level.