0.330 mol of octane is allowed to react with 0.820 mol of oxygen. Which is the limiting reactant?

1 Answer
Apr 9, 2016

Answer:

Dioxygen gas is the limiting reagent. We need the balanced equation representing the combustion of octane:

Explanation:

We need the balanced equation representing the combustion of octane:

#C_8H_20(l) + 13O_2(g) rarr 8CO_2(g) + 10H_2O(g)#

Is this stoichiometrically balanced? Do not trust my arithmetic!

The balanced equation specifies that #1# #"mole"# of octane requires #13# #"moles"# of dioxygen gas for complete combustion. Equivalently, that each #114# #g# of octane requires some #336# #g# dioxygen for complete combustion. From where do I get these numbers?

Most of the time, in internal combustion, and diesel engines, and furnaces we would expect SOME of the hydrocarbon to combust incompletely to give #CO#, and #C# (as soot) as oxidation products. I could represent this reaction by listing carbon and carbon monoxide as combustion products and balancing the equation appropriately:

#C_8H_20(l) + 23/2O_2(g) rarr 6CO_2(g) + CO(g) + C(s) + 10H_2O(g)#

Would I know that I produced the #1# mole of soot and carbon monoxide in the reaction? Of course, I would not. I would have to analyze the flue gases to make a stoichiometric appraisal. Nevertheless, this reaction represents an incomplete combustion.

How would energy transfer in these reactions? Would they produce or consume energy? Why?