How much air will it take to burn 320 grams of methanol?

The oxidation reaction of methanol is described by the following chemical reaction equation.

CH3OH + 3/2 O2 = CO2 + 2H2O;

According to the coefficients of this equation, 1.5 oxygen molecules are required to oxidize 1 molecule of methanol. In this case, 1 molecule of carbon dioxide is synthesized.

Let’s calculate the amount of methanol available.

To do this, we divide the mass of the substance by its molar weight.

M CH3OH = 12 + 4 + 16 = 32 grams / mol;

N CH3OH = 320/32 = 10 mol;

The amount of oxygen will be.

N O2 = 10 x 1.5 = 15 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 = 15 x 22.4 = 336 liters;

Provided that the volume fraction of oxygen in the air is 21%, the required air volume will be 336 / 0.21 = 1600 liters = 1.6 m3;