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September 1, 2014

Homework Help: Chemistry

Posted by Lena on Wednesday, November 5, 2008 at 9:33pm.

I did a lab where I put Copper (II) Chloride in water with Alumnim.

2Al + 3CuCl2 -> 3Cu + 2AlCl3

The aluminum on the reactant side weighs 0.5 g and the CuClw weighs 1.69 grams.

It said to find the theoricital yield. Did I do this right?

2Al + 3CuCl2 -> 3Cu + 2AlCl3


# of mols Al = 0.50 g / 26.98 g/mol = 0.018532246 mol.
# of mols CuCl2 = 1.69 g / [63.55+(35.45 x 2)] g/mol = 0.012569728 mol

(0.012569728 mol CuCl2) * n mol AlCl3 / 0.012569728 mol CuCl2 = (0.012569728 mol CuCl2) * 2 mol AlCl3 / 3 mol AlCl3
n mol AlCl3 = 0.008379818

(0.0185322468 mol Al) * n mol AlCl3 / 0.018532246 mol Al = (0.018532246 mol Al) * 2 mol AlCl3 / 2 mol Al
n mol AlCl3 = 0.018532246

Therefore, the limiting reagent is the Copper (II) Chloride

Theoretical Yield of Aluminium Chloride:

Mass of AlCl3 = (0.008379818 mol) * (133.33 grams/mol)
= 1.1173 grams

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