Question

When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction, 2 ch3oh(

Answer 1

The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH 
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH

Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction