Answer to Question #115177 in General Chemistry for Janet

Question

A 8.1 g piece of aluminum foil is placed in a solution containing 40.03 g of Copper (II) Chloride. What would be the resulting colour of the solution in this scenario? Use stoichiometry calculations to solve this question.

Answer 1

We have the following reaction equation:

2Al + 3CuCl₂ = 3Cu + 2AlCl₃

We have

n(Al) = 8.1/27 = 0.3 mol and n(CuCl₂) = 40.03/135 = 0.296 mol.

According to the stoichiometry of the reaction - aluminum in excess, for the complete interaction of 0.3 mol of copper (II) chloride requires 0.2 mol of aluminum foil. Aluminium will completely displace copper from copper (II) chloride, which will be accompanied by discoloration of the solution. Because aluminum (III) chloride has no color, the final solution is colorless.