Answer to Question #263527 in General Chemistry for Nica

Question

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

Answer 1

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.