How many liters of air will it take to burn 60 grams of ethanol?

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 = 60/46 = 1.304 mol;

The amount of oxygen will be.

N O2 = 1.304 x 3 = 3.912 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 = 3.912 x 22.4 = 87.629 liters;

The required volume of air will be: V air = 87.629 / 0.21 = 417.3 liters;