How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a temperature of 850 c?

First, we need the no.of moles of O2 = mass/molar mass of O2
                                                             = 55 g / 32 g/mol
                                                             = 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2 
So we can get the no.of moles of H2O = 2 * moles of O2
                                                                  = 2 * 1.72 mol
                                                                  = 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm  & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
 ∴ V ≈ 8.2 L