How much oxygen is needed for the complete combustion of 96 g of ethyl alcohol?

The reaction of burning ethyl alcohol is described by the following chemical reaction equation.

C2H5OH + 3O2 = 2CO2 + 3H2O;

When burning 1 mol of ethyl alcohol, 3 mol of oxygen is used.

Let’s calculate the chemical amount of a substance in 96 grams of ethyl alcohol.

M C2H5OH = 12 x 2 + 6 +16 = 46 grams / mol;

N C2H5OH = 96/46 = 2.087 mol;

This will involve 2.087 x 3 = 6.261 mol of oxygen.

Let’s calculate its volume.

Under normal conditions, one liter of ideal gas assumes a volume of 22.4 liters.

V O2 = 6.261 x 22.4 = 140.25 liters;

The volume fraction of oxygen in the air is 20.95%.

Taking this into account, the air volume will be:

V air = 140.25 / 0.2095 = 669.45 liters;