When 12 moles of methanol (ch3oh) and 24 moles of oxygen gas react according to the chemical equation below, what is the limiting reactant and how much of the excess reactant is left over?

The balanced reaction equation for the reaction between CH₃OH and O₂ is 
                        2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles        12                24
Reacted moles   12                18
Final moles          -                   6              12               24
 
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
   reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
                                      = 12 mol x 3 / 2  
                                      = 18 mol

All of CH₃OH moles react with O₂.

Hence, the limiting agent is CH₃OH. 

Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
                                            

Yapo 13 fool coming at you for it