How much air is needed for the complete combustion of 5 L of ethanol (ρ=0,79 g/сm3)? [A: 28,85 m3] C2H5OH + 3 O2 = 2 CO2 + 3 H2O?

5L ethanol = 5000mL = 5000cm³ * 0.78g/cm³ =3950g
Molar mass C2H5OH =46g/mol
Mol ethanol = 3950g /46g/mol = 85.87 mol
This will require 85.87*3 = 257.6mol O2
At STP 1mol O2 has volume = 22.4L
257.6 mol has volume = 5,770 L
Air is 5x volume of O2
Volume of air = 5,770*5 = 28,850L
1000L = 1m³
Volume m³ = 28.85m³

Thank you so much. But Can you check why I can't do this in following way:

According to reaction: C2H5OH + 3 O2 = 2 CO2 + 3 H2O
for combustion 1 mol C2H5OH needs 3 mol O2
that means 22.4 L C2H5OH needs (3*22.4) or 67.2 L O2
so, 5L C2H5OH needs = (67.2 * 5) / 22.4 0r 15 L O2
Air is 5x volume of O2
Volume of air = 15*5 = 75L

I can't understand where misconception is happened here.