How much air is required for the complete combustion of 7.8 g of benzene?

The reaction of burning benzene is described by the following chemical reaction equation.

C6H6 + 7 / 2O2 = 3CO2 + 3H2O;

When burning 1 mol of benzene, 3.5 mol of oxygen is used.

Let’s calculate the chemical amount of a substance in 7.8 grams of benzene.

M C6H6 = 12 x 6 + 6 = 78 grams / mol;

N C6H6 = 7.8 / 78 = 0.1 mol;

In this case, 0.1 x 3.5 = 0.35 mol of oxygen will be involved.

Let’s calculate its volume.

Under normal conditions, one liter of ideal gas assumes a volume of 22.4 liters.

V O2 = 0.35 x 22.4 = 7.84 liters;

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

Taking this into account, the air volume will be:

V air = 7.84 / 0.2095 = 37.4 liters;