What volume of carbon monoxide 4 will be released during the combustion of hexane weighing 500 g

What volume of carbon monoxide 4 will be released during the combustion of hexane weighing 500 g, if the mass fraction of incombustible impurities in this hexane sample is 8%.

The oxidation reaction of hexane with oxygen is described by the following chemical reaction equation:

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

19 moles of oxygen are reacted with 2 moles of hexane. This synthesizes 12 mol of carbon monoxide.

Let’s calculate the available chemical amount of hexane.

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

N C6H14 = 500 x 0.92 / 86 = 5.349 mol;

For this reaction, 5.349 x 12/2 = 32.094 mol of oxygen will be released.

Let’s calculate the volume of oxygen. To do this, we multiply the amount of substance and the standard volume of 1 mole of gaseous substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V O2 = 32.094 x 22.4 = 718.91 liters;