Question

Given the balanced ionic equation: 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s) Compared to the total charge of the reactants, the total charge of the products is 1. less 2. greater 3. the same

Answer 1

3.- The same

Step-by-step explanation:

1.- In the reactants you need to calculate the charges that you have:

2 Al(s) = zero, because the aluminum is in it based formed (that mean without charge).

3 Cu2+ = 3 x 2+ = 6+ That is the total positive charges that copper collaborate in this reaction.

2.- Then calculate the charges on the products:

2 Al3+ = 2 x 3+ = 6+ charges from the aluminum.

3 Cu(s) = zero, because the copper in this case is in the base form

3.- In this way, the charges at the begging (6+) and at the end (6+) are the same.