Can you represent the combustion of pentane by means of a stoichiometric equation?

1 Answer
Feb 21, 2017

Answer:

#C_5H_12(l) +8O_2(g) rarr 5CO_2(g) + 6H_2O(l)#

Explanation:

The given equation is stoichiometrically balanced: garbage in equals garbage out. If the pentane is in stoichiometric proportion with the dioxygen and #"10 mol dioxygen"# were used, then there were #5/4*mol# of #"pentane"# precisely, and #25/4*mol# of carbon dioxide gas would result.

Why? Because I merely follow the stoichiometry of the rxn.
#"1 mole"# or #"1 equiv"# #"pentane"# react with #"8 equiv"# dioxygen to give #"5 equiv"# carbon dioxide, and #"6 equiv"# water. In an internal combustion engine, however, combustion is rarely so complete. The products of incomplete combustion, #CO#, and #C#, are known to occur.