How much oxygen is needed for the complete combustion of 96 g of ethyl alcohol?

The oxidation reaction of ethyl alcohol with oxygen occurs in accordance with the following chemical reaction equation:

C2H5OH + 2O2 = 2CO2 + 3H2O;

To burn 1 mol of ethyl alcohol, 2 mol of oxygen is required.

Find the required amount of ethyl alcohol.

М С2Н5ОН = 12 х2 + 5 + 16 + 1 = 46 grams / mol;

N C2H5OH = 96/46 = 2.087 mol;

To oxidize this amount of ethyl alcohol, 2.087 x 2 = 4.174 mol of oxygen is required.

Let’s find its volume.

One mole of gas under normal conditions fills a volume of 22.4 liters.

The oxygen volume will be equal to:

V O2 = 4.174 x 22.4 = 93.4976 liters;