To burn 624 g of benzene under normal conditions, will you need a volume of oxygen?

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

2C6H6 + 15O2 = 12CO2 + 6H2O;

15 moles of oxygen are reacted with 2 moles of benzene. This synthesizes 12 mol of carbon monoxide.

Let’s calculate the available chemical amount of benzene substance.

M C6H6 = 12 x 6 + 6 = 78 grams / mol;

N C6H6 = 624/78 = 8 mol;

This reaction will require 8 x 15/2 = 60 moles of oxygen.

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 = 60 x 22.4 = 1 344 liters;