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Firstly, you need a balanced equation:

2Al(s) + 3CuCl2 (aq) ----> 2AlCl3 (aq) + 3Cu (s)

Now, you need to convert grams to moles:

Moles of Al = 0.25 g x 26.98 g mol-1 = 9.266 x 10-3 mol Al (s)

From the equation above we see that the ratio of CuCl2 to Al(s) is simply 3:2 so we must work out the moles of CuCl2 required to react completely with Al(s):

Moles of CuCl2 = 9.266 x 10-3 x 3/2 = 0.0139 mol CuCl2

Next convert moles to grams:

Grams CuCl2 = 0.0139 mol CuCl2 x 134.45 g mol-1 = 1.869 g

This means that you would need 1.869g of Copper II Chloride to reacted completely with 0.25 g Aluminium.

Therefore, Copper II Chloride is the Limiting Reactant.

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Q: If you have 0.25 g of aluminium foil and 0.51 g of copper II chloride which is the limiting reactant?
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