. How long would a current of 70.0 amps have to run to produce 25.0 g of aluminum?Al3+(aq) → Al(s)21.3 minutes42.5 minutes63.8 minutes3830 minutes

To solve this problem, we need to use Faraday's law of electrolysis which states that the amount of substance produced at an electrode during electrolysis is directly proportional to the number of moles of electrons (or the amount of electric charge) transferred at that electrode.

The balanced equation for the reduction of aluminum is:

2Al3+ + 6e- → 2Al

From the balanced equation, we can see that 3 moles of electrons are needed to produce 1 mole of aluminum.

First, we need to convert the mass of aluminum to moles. The molar mass of aluminum is approximately 26.98 g/mol.

25.0 g Al * (1 mol Al / 26.98 g Al) = 0.926 moles Al

Since 3 moles of electrons are needed to produce 1 mole of aluminum, we need:

0.926 moles Al * (3 moles e- / 1 mol Al) = 2.778 moles e-

The charge of one mole of electrons (1 Faraday) is approximately 96485 Coulombs. So, the total charge needed is:

2.778 moles e- * (96485 C / 1 mol e-) = 268,000 C

The current (I) is the charge (Q) divided by the time (t), so to find the time, we rearrange the formula to:

t = Q / I

Substituting the given current of 70.0 amps (which is equivalent to 70.0 Coulombs per second), we find:

t = 268,000 C / 70.0 C/s = 3830 seconds

To convert seconds to minutes, divide by 60:

3830 s * (1 min / 60 s) = 63.8 minutes

So, a current of 70.0 amps would have to run for approximately 63.8 minutes to produce 25.0 g of aluminum. Therefore, the correct answer is 63.8 minutes.