Determine the mass of hexane during the combustion of which 72 g of water was released.

Determine the mass of hexane during the combustion of which 72 g of water was released. Calculate the volume of air consumed to burn this mass of hexane.

The hexane oxidation reaction is described by the following chemical reaction equation.

2C6H14 + 19O2 = 12CO2 + 14H2O;

According to the coefficients of this equation, 19 oxygen molecules are required to oxidize 2 hexane molecules. In this case, 12 molecules of carbon dioxide and 14 molecules of water are synthesized.

Let’s calculate the synthesized amount of water.

To do this, divide the weight of the resulting water by the weight of 1 mole of water.

M H2O = 2 + 16 = 18 grams / mol;

N H2O = 72/18 = 4 mol;

Let’s calculate the amount of hexane that reacted.

N C6H14 = 4/14 x 2 = 0.571 mol;

Its weight will be.

M C6H14 = 12 x 6 + 14 = 86 grams / mol;

m C6H14 = 0.571 x 86 = 49.1 grams;

The amount of oxygen will be.

N O2 = 4/14 x 19 = 5.429 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 = 5.429 x 22.4 = 121.6 liters;

The oxygen content in the air is 21%.

The required air volume will be:

V air = 121.6 / 0.21 = 579.1 liters;