What volume of oxygen and air is required for complete combustion of 23 grams of ethanol

What volume of oxygen and air is required for complete combustion of 23 grams of ethanol, the volume fraction of oxygen in the air is 21%?

The ethanol oxidation reaction is described by the following chemical reaction equation.

C2H5OH + 3O2 = 2CO2 + 3H2O;

According to the coefficients of this equation, for the oxidation of 1 molecule of alcohol, 3 molecules of oxygen are required. In this case, 2 molecules of carbon dioxide are synthesized.

Let’s calculate the amount of ethanol available.

To do this, we divide the mass of the substance by its molar weight.

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

N C2H5OH = 23/46 = 0.5 mol;

The amount of oxygen will be.

N O2 = 0.5 x 3 = 1.5 mol;

Let’s calculate the gas volume. To do this, multiply the amount of substance by the volume of 1 mole of gas.

Its volume will be: V O2 = 1.5 x 22.4 = 33.6 liters;

The required volume of air will be: V air = 33.6 / 0.21 = 160 liters;