Set the volume of air that is required for burning ethyl alcohol with a mass of 92g.

The combustion reaction of ethyl alcohol is described by the following chemical reaction equation:

C2H5OH + 3O2 = 2CO2 + 3H2O;

For the combustion of one mole of ethyl alcohol, three moles of oxygen are required.

Determine the amount of substance in 92 grams of ethyl alcohol.

M C2H5OH = 12 x 2 + 5 + 16 + 1 = 46 grams / mol;

N C2H5OH = 92/46 = 2 mol;

Determine the required air volume.

N O2 = N C2H5OH x 3 = 2 x 3 = 6 mol;

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

V O2 = 6 x 22.4 = 134.4 liters;

The volume fraction of oxygen in the air is 20.9476%;

V air = V O2 / 0.209476 = 134.4 / 0.209476 = 641.6 liters;