Determine the amount of air that will be consumed to burn 4,6 ethanol.

The ethanol combustion reaction is described by the following chemical equation:

C2H5OH + O2 = 2CO2 + 3H2O;

1 molecule of alcohol interacts with one molecule of oxygen.

Let’s calculate the chemical amount of a substance in 4.6 grams of ethanol.

Its molar mass is:

M C2H5OH = 12 x 2 + 6 + 16 = 46.0 grams / mol;

N C2H5OH = 4.6 / 46.0 = 0.1 mol;

The same amount of oxygen will be required.

1 mole of ideal gas normally takes on a volume of 22.40 liters.

The oxygen volume will be:

V O2 = 0.1 x 22.40 = 2.240 liters;

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

The required air volume is:

V air = V O2 / 0.2095 = 2.24 / 0.2095 = 10.69 liters;