Calculate the volume of oxygen required to burn 4.6 g of ethanol.

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

C2H5OH + 3O2 = 2CO2 + 3H2O;

3 moles of oxygen interact with 1 mole of glycerol. This synthesizes 2 mol of carbon monoxide.

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

M C2H5OH = 12 x 2 + 6 + 16 = 46 grams / mol;

N C2H5OH = 4.6 / 46 = 0.1 mol;

This reaction will require 0.1 x 3 = 0.3 mol of oxygen.

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

V O2 = 0.3 x 22.4 = 6.72 liters;

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