Magnesium reacts with aluminum chloride. if you have 40g of each, how many grams of al can you make? what is the lr? how much excess is left over after the reaction takes place

Q1)
Magnesium reacts with AlCl₃ and displaces Al.
3Mg + 2AlCl₃ ---> 3MgCl₂ + 2Al
stoichiometry of Mg to AlCl₃ is 3:2
Mass of Al₂Cl₃ present - 40 g
Therefore number of Al moles - 40 g/ 133.3 g/mol = 0.3 mol
Mass of Mg present - 40 g
Therefore number of moles - 40 g/ 24.3 g/mol = 1.6 mol
If Mg is the limiting reactant;
number of AlCl₃ moles that react with 3 mol of Mg  - 2 mol
Therefore number of AlCl₃ moles with 1.6 mol - 2/3 x 1.6 = 1.07 mol
However only 0.3 mol of Al present, therefore Mg is in excess and AlCl₃ is the limiting reactant .

Q2)
stoichiometry of AlCl₃ to Al is 2:2 
SInce AlCl₃ is the limiting reactant, the amount of Al produced depends on the amount of limiting reactant present.
Number of AlCl₃ moles reacted - 0.3 mol
Since molar ratio of AlCl₃ to Al is 1:1
the number of Al moles produced - 0.3 mol
The mass of Al produced - 0.3 mol x 27 g/mol = 8.1 g

Q3) AlCl₃ is the limiting reactant. limiting reactant is the reagent which is fully used up in the reaction. Reagent in excess is the reactant which is present in excess, amount present is more than what is required for the reaction.
Number of AlCl₃ moles reacted - 0.3 mol
Number of Mg moles reacted with 2 mol of AlCl₃ - 3 mol
Therefore with 0.3  mol of AlCl₃ - 3/2 x 0.3 mol = 0.45 mol
Number of Mg moles present initially - 1.6 mol 
moles reacted - 0.45 mol
therefore excess moles left after the reaction - 1.6 - 0.45 = 1.15 mol

What record? the question is incomplete