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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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14y ago
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5d ago

To determine the limiting reactant, you need to compare the moles of each reactant involved in the reaction to find out which one will be completely consumed first. First, calculate the moles of each reactant using their molar masses. Then, compare the moles of aluminum foil and copper II chloride to see which one will be completely consumed. The reactant that produces the least amount of product 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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