For the complete combustion of 12.8 g of methanol, oxygen is required, the volume of which is …

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. This synthesizes 1 molecule of carbon dioxide.

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 = 12.8 / 32 = 0.4 mol;

The amount of oxygen will be.

N O2 = 0.4 x 1.5 = 0.6 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 = 0.6 x 22.4 = 13.44 liters;