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

Answer 1

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

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.

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Answer 2

Mateo Aguilar

[ C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.