Determine the volume of air (oxygen 21%) required for the combustion of ethyl alcohol with a volume of 500 ml

Determine the volume of air (oxygen 21%) required for the combustion of ethyl alcohol with a volume of 500 ml (density = 0.8 g / ml) and the mass fraction of C2H5OH 92%

Given: W (O2) = 21% = 0.21
V (C2H5OH) = 500ml
p = 0.8g / ml
W (C2H5OH) = 92% = 0.92
Find: Vair-?
The equation:
С2Н5ОН + 3О2 = 2СО2 + 3Н2О
Decision:
m (C2H5OH) = 500 * 0.8 = 400 grams
m (C2H5OH pure) = 400 * 0.92 = 368 grams
M (C2H5OH) = 46gram / mol
n = 368/46 = 8 mol.
n (O2) = 8 * 3 = 24 mol.
The molar volume is 22.4 liters
V (pure O2) = 22.4 * 24 = 537.6 liters.
V (air) = 537.6 / 0.21 = 2560 liters = 2.56 cubic meters of air
Answer: 2.56 cubic meters of air.