Step 1
The reaction must be written and balanced:
2Al + 3Cl2 2 AlCl3
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Step 2
Information provided: 25.0 moles Cl2
Information needed: the molar mass of AlCl3 => 133 g/mol
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Step 3
By stoichiometry
1 mole AlCl3 = 133 g/mol
Procedure:
3 x 1 mole Cl2 --------- 2 x 133 g AlCl3
25.0 moles Cl2 --------- X
X = 25.0 moles Cl2 x 2 x 133 g AlCl3/3 x 1 mole Cl2 = 2217 g AlCl3
Answer: 2217 g AlCl3