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

1 Answer
Feb 21, 2017

C_5H_12(l) +8O_2(g) rarr 5CO_2(g) + 6H_2O(l)C5H12(l)+8O2(g)5CO2(g)+6H2O(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"10 mol dioxygen were used, then there were 5/4*mol54mol of "pentane"pentane precisely, and 25/4*mol254mol of carbon dioxide gas would result.

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