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.
Express your answer numerically in litres.
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
Well, if we assume a 1:1 stoichiometry between methane and methanol, then we can say that every mole of methane will produce one mole of methanol.
To find the volume of methanol formed, we need to know the number of moles of methane. So, Let's start by converting the pressure of methane to atm using the conversion factor: 1.013 bar = 1 atm.
Next, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Since we know the pressure, volume, and temperature of methane, and we can assume 1 mole of methane and 1 mole of methanol is formed, we can solve for the volume of methanol using the equation:
V = (nRT)/P
Given that R = 0.0821 L·atm/(mol·K) and T = 25 + 273 = 298 K, we can calculate the volume of methanol using the equation:
V = (1 × 0.0821 × 298)/1
V ≈ 24.36 L
So, approximately 24.36 liters of methanol is formed. Keep in mind that this is just an estimate and the actual volume may vary.