Question: This equation shows how much AlCl3 can be produced with 22.0g of Al. 22.0g Al x (1 mol Al/26.98g) x (2 mol AlCl3/2mol Al) x

This equation shows how much AlCl3 can be produced with 22.0g of Al.

22.0g Al x (1 mol Al/26.98g) x (2 mol AlCl3/2mol Al) x (133.34 g/1 mol AlCl3) = 108.66g AlCl3

This equation shows how much AlCl3 can be produced with 27.0g of Cl2.

27.0g Cl2 x (1 mol Cl2/70.91g) x (2mol AlCl3/3 mol Cl2) x (133.34g/1 mol AlCl3) = 33.87g AlCl3

This shows that chlorine is the limiting reactant. Only 33.87 grams of AlCl3 can be produced before the chlorine will run out.

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