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