How much oxygen is needed to oxidize 38 liters of carbon monoxide (II)?

The oxidation reaction of carbon monoxide is described by the following chemical reaction equation:
CO + ½ O2 = CO2;
1 mole of carbon monoxide reacts with ½ mole of oxygen. This synthesizes 1 mol of carbon dioxide.
Find the chemical amount of a substance contained in 38 liters of carbon monoxide.
To do this, we divide the volume of gas by the volume of 1 mole of gas.
N CO = 38 / 22.4 = 1.696 mol;
To oxidize such an amount of a substance, you need 1.696 / 2 = 0.848 mol of oxygen.
Let’s define its volume.
To do this, multiply the amount of substance by the volume of 1 mole of gas (22.4 liters).
V O2 = 0.848 x 22.4 = 18.995 liters;