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 = 50/46 = 1.087 mol;
The amount of oxygen will be.
N O2 = 1.087 x 3 = 3.261 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 = 3.261 x 22.4 = 73.046 liters;
The required volume of air will be: V air = 73.046 / 0.21 = 347.84 liters;
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