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
The exact answer is 14.85724 grams. Just simply multiply these two numbers together to get your answer in grams. No fancy formula is needed for this! Hopefully this helps!
