How do we represent the combustion of ethanol stoichiometrically?

1 Answer
Jan 31, 2017

Answer:

Balance mass; balance charge.

#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#

Explanation:

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.