Question

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solution containing 0.75 grams of copper (II) chloride. A single replacement reaction takes place. Which statement explains the maximum amount of copper that the chemist can extract using this reaction

Answer 1

Answer & Explanation:

The replacement of aluminium (Al) by copper (Cu) can be represented by the following chemical equation:

3CuCl2 + 2Al  = 2AlCl3 + 3Cu

We are given 0.50g of Al, and 0.75g of CuCl2

We calculate the maximum amount of copper by assuming one of the reactants (CuCl2 or Al) will be depleted.

Calculations:

The corresponding molecular masses are :

3CuCl2 = 3 (63.546+2*35.457) = 403.38 ...........(1)

2Al = 2*26.9815 = 53.963........................................(2)

2AlCl3 = 2 (26.9815+3*35.457) = 266.705  (we don't need to know this!)

3Cu = 3*63.546 = 190.638.......................................(3)

Assuming 0.50g of aluminium will be depleted with unlimited CuCl2, the amount of copper recuperated will be:

M1 = 0.50 *( (3)/(2) ) = 0.50*190.638/53.963 = 1.766 g

If 0.75g of CuCl2 were depleted with unlimited Al, the amount of copper recuperated will be

M2 = 0.75 * ( (3) / (1) ) = 0.75 * 190.638 / 403.38 = 0.354 g   < M1

Therefore CuCl2 is the limiting reagent.

Since we can obtain the maximum amount of copper from the limiting reagent (CuCl2), so the required maximum amount is 0.35 g of copper.