Calculate the volume of air required for the complete combustion of 6.4 g methanol.

The oxidation reaction of methanol with oxygen is described by the following chemical reaction equation:

CH3OH + 2O2 = CO2 + 2H2O;

1 mol of alcohol reacts with 2 mol of oxygen.

Let’s calculate the available chemical amount of the alcohol substance.

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

N CH3OH = 6.4 / 32 = 0.2 mol;

To burn such an amount of matter, you need to take 0.2 x 2 = 0.4 mol of oxygen.

Let’s calculate its volume.

V O2 = 0.4 x 22.4 = 50 liters;

The oxygen content in the air is 21%. The required air volume is:

V air = 50 / 0.21 = 10.5 liters;