How much air will it take to burn 120 g of carbon?

The reaction of carbon with oxygen takes place according to the following chemical reaction equation:

C + O2 = CO2;

For oxidation of 1 mole of carbon, 1 mole of oxygen is required.

Let’s find the amount of carbon.

M C = 12 grams / mol;

N C = 120/12 = 10 mol;

To burn this amount of carbon, 10 moles of oxygen are 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 = 10 x 22.4 = 224 liters;

The volume concentration of oxygen in the air is 20.95%.

The required air volume will be:

V air = 224 / 0.2095 = 1069 liters = 1.069 m3;