How do we represent the combustion of ethanol stoichiometrically?

Question

Peyton Adams · Accepted Answer

Balance mass; balance charge.

#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O# All hydrocarbons are known to combust to give carbon dioxide and water. So we write the equation accordingly: #C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O# Are charge and mass balanced? For every reactant particle, is there a corresponding reactant particle? If there is not, there should be, and you know that you have furrher work to do. The typical order of operations is: balance the carbons; then balance the hydrogens; and lastly balance the oxygens...... #C_6H_5OH + O_2 rarr 6CO_2 + H_2O#; #"carbons balanced."# #C_6H_5OH +O_2 rarr 6CO_2 + 3H_2O#; #"hydrogens balanced."# #C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#; #"oxygens balanced."# The reaction is now stoichiometrically balanced. Will this reaction be exothermic? How do you represent the combustion of ethanol, #C_2H_5OH#, of hexane, #C_6H_14#? The hexane combustion is a trickier proposition on the basis of arithmetic. Try it out. Especially with aromatics, sometimes incomplete combustion occurs to give #C#, as soot, and #CO# as combustion products. For more of the same see here.

Wyatt Atkins · Answer

undefined The stoichiometric combustion of ethanol (C2H5OH) can be represented by the following balanced chemical equation:

C2H5OH + 3O2 → 2CO2 + 3H2O