Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2al(s)+3cl2(g)→2alcl3(s) what is the maximum mass of aluminum chloride that can be formed when reacting 21.0 g of aluminum with 26.0 g of chlorine?

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

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