2Al (s) + 3Cl2 (g)  2 AlCl3 (s)

Question

i)obtain the limiting reagent.
when 2.00 mole of Cl2 (g) is reacted with 54.3 g of Al (p).
ANSWER : Cl is limiting reagant.
54.3gAl(p)/27 = 2.01AI(P)
Initial = 2.01 mol 2 mol
divied = 2.01/2 2/3
= 1.005 =0.67
cl is smaller than al .
can some one check my answer because i confuse with this question.

Answer 1

No. Al is the limiting reagent.
(and I don't understand what (p) is). And I don't understand your work either for apparently you have omitted division signs and multiplications signs.

54.3/27 = 2.01 is ok. Convert that to mols Cl2 needed. That's 2.01 mols Al x (3 mols Cl2/2 mols Al) = 3.01 mols Cl2 needed and you don't have that much Cl2 so Al is the limiting reagent.

Or you could have started with Cl2. You have 2 mols Cl2. How much Al is needed to react with that 2 mols Cl2. 2.0 mols Cl2 x (2 mols Al/3 mols Cl2) = 1.33 mols Al needed and you have that much so Al is the LR.