How much oxygen is consumed to burn 80 grams of butanol?

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

C4H9OH + 6O2 = 4CO2 + 5H2O |;

According to the coefficients of this equation, 6 oxygen molecules are required to oxidize 1 butanol molecule. In this case, 4 molecules of carbon dioxide are synthesized.

Let’s calculate the amount of butanol available.

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

M C4H9OH = 12 x 4 + 10 + 16 = 74 grams / mol;

N C4H9OH = 80/74 = 1.081 mol;

The amount of oxygen will be.

N O2 = 1.081 x 6 = 6.486 mol;

Let’s calculate the gas volume. To do this, we multiply the amount of substance by the volume of 1 mole of gas.

Its volume will be: V O2 = 6.486 x 22.4 = 145.29 liters;