2Al (s) + 3Cl2 (g)  2 AlCl3 (s)

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.

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.

To determine the limiting reactant, we need to compare the moles of Cl2 to the moles of Al.

Given:
Moles of Cl2 = 2.00 mol
Mass of Al = 54.3 g

To find the moles of Al, we need to convert the mass of Al to moles.

Molar mass of Al = 27 g/mol

Moles of Al = mass of Al / molar mass of Al
= 54.3 g / 27 g/mol
= 2.01 mol

Now, let's compare the moles of Cl2 and Al:
Moles of Cl2: 2.00 mol
Moles of Al: 2.01 mol

Based on the stoichiometry of the balanced chemical equation, the ratio between Cl2 and Al is 3:2.

From the calculation, we can see that the moles of Cl2 (2.00 mol) are greater than the moles of Al (2.01 mol).

Thus, the limiting reactant is Al (s) because it will be completely consumed in the reaction, and there will be excess Cl2 left over.

Therefore, the given answer "Cl is limiting reagent" is incorrect.

To determine the limiting reagent in this reaction, you need to compare the number of moles of each reactant and see which one is present in a lesser amount.

First, convert the mass of aluminum (Al) to moles:
mass of Al = 54.3 g
molar mass of Al = 27 g/mol
moles of Al = mass of Al / molar mass of Al = 54.3 g / 27 g/mol = 2.01 mol

Now, compare the number of moles of Cl2 with that of Al:
moles of Cl2 = 2.00 mol (given)

From the balanced chemical equation, we can see that the stoichiometry between Al and Cl2 is 2:3. This means that 2 moles of Al react with 3 moles of Cl2.

Now we calculate the moles of Al for 2 moles of Cl2:
moles of Al = (2.00 mol Cl2) * (2 mol Al / 3 mol Cl2) = 1.33 mol Al

Since we have 2.01 moles of Al and only 1.33 moles of Al are required to react with 2.00 moles of Cl2, Al is present in excess.

Therefore, the limiting reagent is Cl2.

It appears that your answer is incorrect. After calculating, it is clear that Cl2 is the limiting reagent, not Al as you initially mentioned.