What mass of aluminum will be produced by the electrolysis of molten AlCl3, if a current of 50.0 amps is run for 30.0 minutes?8.40 g Al25.2 g Al0.14 g Al75.6 g Al

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 transferred at that electrode.

Step 1: Convert the time from minutes to seconds because the unit of current (ampere) is coulombs per second. 30.0 minutes = 30.0 * 60 = 1800 seconds

Step 2: Calculate the total charge (in coulombs) that passed through the solution. Total charge = current * time = 50.0 amps * 1800 seconds = 90000 coulombs

Step 3: Calculate the number of moles of electrons that were transferred. We know that 1 mole of electrons carries a charge of approximately 96485 coulombs (Faraday's constant). Number of moles of electrons = total charge / Faraday's constant = 90000 coulombs / 96485 coulombs/mole = 0.933 moles of electrons

Step 4: Determine the stoichiometry of the reaction. The reaction for the electrolysis of AlCl3 is: 2 AlCl3 + 3 e- -> 2 Al + 3 Cl2 From the reaction, we can see that 3 moles of electrons produce 2 moles of aluminum.

Step 5: Calculate the number of moles of aluminum produced. Number of moles of Al = (0.933 moles of electrons) * (2 moles of Al / 3 moles of electrons) = 0.622 moles of Al

Step 6: Convert the number of moles of aluminum to grams using the molar mass of aluminum (26.98 g/mole). Mass of Al = number of moles of Al * molar mass of Al = 0.622 moles of Al * 26.98 g/mole = 16.8 g of Al

So, none of the given options is correct. The mass of aluminum produced would be approximately 16.8 g.