# What volume of air is required for complete combustion of 0.5 kg of butanal?

What volume of air is required for complete combustion of 0.5 kg of butanal? The volume fraction of oxygen in the air is 21%.

The oxidation reaction of butanal is described by the following chemical reaction equation.

C4H8O + 11 / 2O2 = 4CO2 + 4H2O;

According to the coefficients of this equation, 5.5 oxygen molecules are required to oxidize 1 butanal molecule. In this case, 4 molecules of carbon dioxide are synthesized.

Let’s calculate the available amount of butanal.

To do this, divide the weight of the available gas by the weight of 1 mole of this gas.

М С4Н8О = 12 x 4 + 8 + 16 = 72 grams / mol;

N C4H8O = 500/72 = 6.944 mol;

The amount of oxygen will be.

N O2 = 6.944 x 11/2 = 38.192 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 = 38.192 x 22.4 = 855.5 liters;

The oxygen content in the air is 21%.

The required air volume will be:

V air = 855.5 / 0.21 = 4073.8 liters = 4.07 m3; One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.