How much air is required for the complete combustion of 10 g of acetone?

The acetone combustion reaction is described by the following chemical reaction equation.

C3H6O + 4O2 = 3CO2 + 3H2O;

When burning 1 mol of acetone, 4 mol of oxygen is used.

Let’s calculate the chemical amount of a substance in 10 grams of acetone.

M С3Н6О = 12 x 3 + 6 +16 = 58 grams / mol;

N C3H6O = 10/58 = 0.172 mol;

This will involve 0.172 x 4 = 0.688 mol of oxygen.

Let’s calculate its volume.

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

V O2 = 0.688 x 22.4 = 15.41 liters;

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

Taking this into account, the air volume will be:

V air = 15.41 / 0.2095 = 73.56 liters;