How many hours are required for the deposition of 15.0 grams of Al from molten AlCl3 by a current of 15.0 amps?
It will require 96,485 coulombs to deposit 27/3 = 9 g Al. Therefore, it will require 96,485 x (15/9) coulombs to deposit 15 g.
1 C = Amp x sec. You know C and amps solve for seconds.
To determine the time required for the deposition of 15.0 grams of Al from molten AlCl3 using a current of 15.0 amps, you need to use the Faraday's Laws of Electrolysis.
Faraday's first law states that the amount of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
The equation that represents the relationship is:
Q = n * F
Where:
Q is the quantity of electricity (coulombs),
n is the number of moles of the substance, and
F is the Faraday's constant, which is equal to 96,485 C/mol.
In this case, you want to know the time required, which can be calculated by rearranging the equation and solving for time (t):
t = Q / I
Where:
t is the time (in seconds),
Q is the quantity of electricity (coulombs), and
I is the current (in amps).
Now, let's calculate the time required:
Step 1: Convert the mass of Al to moles using the molar mass of Al.
Molar mass of Al = 26.98 grams/mol
Number of moles of Al = 15.0 grams / 26.98 grams/mol
Step 2: Calculate the quantity of electricity (Q) using Faraday's law.
Q = n * F = (number of moles of Al) * 96,485 C/mol
Step 3: Calculate the time required (t) using the equation:
t = Q / I
Substitute the values into the equation:
t = (Q) / (I)
t = [(number of moles of Al) * 96,485 C/mol] / (15.0 amps)
Now you can plug in the values to find the answer.