How much oxygen is required to burn 800 g of ethyl alcohol containing 40% impurity?

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

C2H5OH + 3O2 —> 2CO2 + 3H2O;

Let’s determine the chemical amount of alcohol. To do this, divide its weight by the weight of 1 mole of the substance.

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

N C2H5OH = 800 x 0.6 / 46 = 10.435 mol;

With this amount of alcohol, 10.435 x 3 = 31.3 mol of oxygen will react.

Let’s calculate its volume.

To do this, multiply the amount of the substance by the volume of 1 mole of the substance (which is 22.4 liters).

V O2 = 31.3 x 22.4 = 701.2 liters;