Respuesta :

Answer:

524.82 g Cl2

Explanation:

Balanced Equation:

2Al(s) + 3Cl(g) ⇒ 2AlCl (s)

Al: 26.982 g/mol     Cl: 35.453 g/mol

266.28g Al x 1mol Al/ 26.982g Al x 3mol Cl2/ 2mol Al x

35.453g Cl2/ 1mol Cl2 = 524.82 g Cl2