The balanced equation for the reaction of aluminum metal and chlorine gas is

Question

2Al(s) + 3Cl2(g) → 2AlCl3(s)

Assume that 0.40 g Al is mixed with 0.25 g Cl2.
What is the maximum amount of AlCl3, in grams, that can be produced?

Answer 1

This is a limiting reagent (LR) problem and is worked like two stochiometry problems.
2Al(s) + 3Cl2(g) → 2AlCl3(s)
mols Al = 0.4/about 27 = approx 0.015 but that's an estimate. You should use better numbers throughout the problem because all of my numbers are estimates.
How many mols COULD that produce? That's
0.015 mols Al x (2 mols AlCl3/2 mols Al) = 0.015 mols Al
mols Cl2 gas = 0.25/71 = est 0.0035 mols. How many mols AlCl3 can we get from this. That's 0.0035 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = est 0.002 mols AlCl3.
So you have two choices. You might get 0.002 mols AlCl3 OR you could get 0.015. Obviously you can get only the smaller amount which means you can produce about 0.002 mols AlC3 and Cl2 is the LR. Redo the numbers to get a more accurate figure and convert mols to grams. Post your work if you get stuck.