Answer:
32.6 g AlCl₃
Explanation:
First, the limiting reagent must be found by determining the moles of aluminum and the moles of chlorine. The atomic/molecular weights of Al and Cl₂ are 26.9815 g/mol and 70.9060 g/mol, respectively.
Al: (21.0 g) / (26.9815 g/mol) = 0.778 mol Al
Cl₂: (26.0g) (70.9060 g/mol) = 0.367 mol Cl₂
The amount of Cl₂ required to react with 0.778 mol Al is 1.167 mol, based on the molar ratio (see below). Therefore, Cl₂ is the limiting reagent.
(0.778 mol Al) x (3Cl₂/2Al) = 1.167 mol Cl₂
The amount of aluminum chloride formed is based on the amount of limiting reagent available:
(0.367 mol Cl₂)(2AlCl₃/3Cl₂) = 0.2446 mol AlCl₃
Finally, the mass of AlCl₃ can be calculated using the molecular weight (133.34 g/mol):
(0.2446 mol)(133.34 g/mol) = 32.6 g AlCl₃
