What volume of methanol is formed if 3.03×1011 L of methane at 1.013 bar pressure and 25 ∘C is oxidized to methanol? The density of CH3OH is 0.791 g mL−1 . Assume that the oxidation of methane to methanol occurs in a 1:1 stoichiometry.

Question

Express your answer numerically in litres.

Answer 1

I assume that is a density of 0.791 g/mL = 791 g/L @ 25C. This may not be the shortest way to work the problem.
CH4 ---> CH3OH

PV = nRT and n = PV/RT
P = 1.013 bar = 1 atm
V = 3.03E11 L @ 25 C (298 K)
n = 1*3.03E11 L/0.08206*298 = 0.1239E11 moles CH4
That will produce 0.1239E11 moles CH3OH @ 25 C and 1.013 bar P since it is a 1:1 stoichiometry.
0.1239E11 mols CH3OH x 32 = 3.965E11 grams CH3OH @ 25 C and 1.013 bar P.
volume = mass/density = 3.965E11 g/791 g/L = 0.005013E11L = 5.013E11 L