What volume of air is required for the oxidation of methanol if you need to obtain 1 ton of solution

What volume of air is required for the oxidation of methanol if you need to obtain 1 ton of solution with a methanal mass fraction of 40%?

Given:
m solution (CH2O) = 1 t = 1,000,000 g
ω (CH2O) = 40%

Find:
V (air) -?

1) 2CH3OH + O2 => 2CH2O + 2H2O;
2) m (CH2O) = ω * m solution / 100% = 40% * 1,000,000 / 100% = 400,000 mol;
3) M (CH2O) = Mr (CH2O) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 1 + 1 * 2 + 16 * 1 = 30 g / mol;
4) n (CH2O) = m / M = 400000/30 = 13333.33 mol;
5) n (O2) = n (CH2O) / 2 = 13333.33 / 2 = 6666.67 mol;
6) V (O2) = n * Vm = 6666.67 * 22.4 = 149333.41 l;
7) V (air) = V (O2) * 100% / φ (O2) = 149333.41 * 100% / 21% = 711111.48 l = 711.1 m3.

Answer: The air volume is 711.1 m3.