How much oxygen is needed to burn 23.4 g of benzene?

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.

To do this, divide the available weight of benzene by the weight of one mole of benzene.

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

N C6H6 = 23.4 / 78 = 0.3 mol;

This reaction will require 0.3 x 15/2 = 2.25 mol of oxygen.

We calculate the volume of oxygen by multiplying the amount of oxygen by the volume of one mole:

V O2 = 2.25 x 22.4 = 50.4 liters;

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