# 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;