When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction, 2 ch3oh(g) + 3 o2(g) ? 2 co2(g) + 4 h2o(g) what is the excess reactant and how many moles of it remain at the end of the reaction?

To determine the amount of excess reactant left, we need to first identify which is the limiting and the excess reactant. We do as follows:

2 CH3OH(g) + 3 O2(g) = 2CO2(g) + 4 H2O(g)

24 mol CH3OH ( 3 mol O2 / 2 mol CH3OH) = 36 mol O2 needed
15 mol O2 ( 2 mol CH3OH / 3 mol O2) = 10 mol CH3OH needed

Since it is oxygen that is used up completely in the reaction, then it is the limiting reactant and the excess reactant would be the methanol. There would be 24 - 10 = 14 mol of methanol excess.