When 2.5g of Al reacts with 5.0g of Cl2, 8.3g of AlCl3 is formed. Calculate the limiting reactants, theoretical yield, and percent yield for the reaction. 2Al + 3Cl3 ---> 2AlCl3

Md Ariful Islam

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2Al + 3Cl2 ==> 2AlCl3

Find limiting reactant:

moles Al present = 12.8 g Al x 1 mol Al/27.0 g = 0.474 moles Al (÷2 ->0.237)

moles Cl2 present = 31.9 g Cl2 x 1 mol/71 g = 0.449 moles Cl2 (÷3 ->0.149) LIMITING REACTANT

Name of limiting reactant is chlorine

mass of AlCl3 formed = 0.449 mol Cl2 x 2 mol AlCl3 / 3 mol Cl2 x 133 g AlCl3/mol = 39.8 g AlCl3

0.551 moles Cl2 available x (2 moles Al/3 moles Cl2) = 0.316 moles Al.

Remaining moles of Al = (0.474 - 0.316) = 0.158 moles Al remaining.

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2Al + 3Cl2 = 2AlCl3

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