0.200 mol of octane is allowed to react with 0.690 mol of oxygen, which is the limiting reactant? 2 C8H18 + 25 O2 ---- 16 CO2 + 18 H2O.

Question

0.200 mol of octane is allowed to react with 0.690 mol of oxygen, which is the limiting reactant? 2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O.

Wyatt Atkins · Accepted Answer

It should be oxygen.

First, we examine the equation: #2C_8H_18(l)+25O_2(g)->16CO_2(g)uarr+18H_2O(l)+Delta# And so, we clearly see that #2# moles of octane require #25# moles of oxygen gas to burn. Since we have #0.2# moles of octane, we require: #0.2color(red)cancelcolor(black)("mol" \ C_8H_18)*(25 \ "mol" \ O_2)/(2color(red)cancelcolor(black)("mol" \ C_8H_18))=2.5 \ "mol" \ O_2# But since we only have #0.69 \ "mol" \ O_2#, we clearly see that it is the limiting reactant in this case.

Natalie Anton · Answer

Well let us rebalance the equation to make our ideas of stoichiometric equivalence a bit more straightforward...

We burn octane and depict full combustion in the conventional manner, which is to balance the carbons as carbon dioxide, the hydrogens as water, and the oxygens last. #C_8H_18(l) + 25/2O_2(g) rarr8CO_2(g) + 9H_2O(l)# #"Moles of octane"=0.200*mol#; and this requires #12*1/2*"equiv"# dioxygen...i.e. #2.50*mol#.... In the reaction dioxygen is thus the LIMITING REAGENT... And here CLEARLY, given that dioxygen is specified to be in #0.690*mol# quantity....octane is in stoichiometric EXCESS...and the reaction would not proceed as written. We COULD work out precisely how much carbon dioxide would result in the case of complete combustion. More realistically, we should be aware that INCOMPLETE combustion to #CO(g)# and #C(s)# would occur, as it certainly does in the internal combustion, and diesel engines. And as a representation we could write... #C_8H_18(l) + 11O_2(g) rarr6CO_2(g)+CO(g) + C(s) + 9H_2O(l)# The limiting reagent for this PROPOSED stoichiometry is once more dioxygen. In summary, the stoichiometry of this reaction is completely out of our control, and the question was posed incorrectly.

Cameron Andrews · Answer

undefined Oxygen is the limiting reactant.

0.200 mol of octane is allowed to react with 0.690 mol of oxygen, which is the limiting reactant? 2 C8H18 + 25 O2 ----&gt; 16 CO2 + 18 H2O.

0.200 mol of octane is allowed to react with 0.690 mol of oxygen, which is the limiting reactant? 2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O.