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 Clâ‚‚ are 26.9815 g/mol and 70.9060 g/mol, respectively.
Al: (21.0 g) / (26.9815 g/mol) = 0.778 mol Al
Clâ‚‚: (26.0g) (70.9060 g/mol) = 0.367 mol Clâ‚‚
The amount of Clâ‚‚ required to react with 0.778 mol Al is 1.167 mol, based on the molar ratio (see below). Therefore, Clâ‚‚ is the limiting reagent.
(0.778 mol Al) x (3Clâ‚‚/2Al) = 1.167 mol Clâ‚‚
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
