How do we represent the combustion of ethanol stoichiometrically?

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.