Question

A chemist wants to extract copper metal from copper chloride solution. The chemist places 1.50 grams of aluminum foil in a solution of 14 grams of copper (II) chloride. A single replacement reaction takes place. What best explains the state of the reaction mixture after the reaction?
Unbalanced equation: CuCl2 + Al → AlCl3 + Cu (4 points)

Group of answer choices

Less than 6.0 grams of copper is formed, and some aluminum is left in the reaction mixture.

More than 6.5 grams of copper is formed, and some aluminum is left in the reaction mixture.

Less than 6.0 grams of copper is formed, and some copper chloride is left in the reaction mixture.

More than 6.5 grams of copper is formed, and some copper chloride is left in the reaction mixture.

Answer 1

Less than 6.0 grams of copper is formed, and some aluminum is left in the reaction mixture

Stoichiometric calculations

From the equation of the reaction:

3CuCl2 + 2Al → 2AlCl3 + 3Cu

Mole ratio of CuCl2 and Al = 3:2

Mole of 1.5 g aluminum foil = 1.5/26.98

                                                    = 0.06 moles

Mole of 14 g CuCl2 = 14/134.45

                                    = 0.10 moles

For every 3 moles of CuCl2, 2 mole of Al is needed.

Thus, for 0.10 moles of CuCl2, the mole of Al needed would be:

                                      0.10 x 2/3 = 0.07 moles

CuCl2 is, thus, slightly in excess while Al is the limiting reagent.

Mole ratio of Al and Cu = 2:3

For 0.06 mole Al, the mole of Cu will be: 0.06x3/2 = 0.09 moles

Mass of 0.09 moles Cu = 0.09 x 63.55

                                            = 5.71 g

Thus, less than 6.0 grams Cu is formed while a little CuCl2 would be left in the reaction mixture.

More on stoichiometric calculations can be found here: brainly.com/question/8062886