How many hours are required for the deposition of 15.0 grams of Al from molten AlCl3 by a current of 15.0 amps?
Chemistry - DrBob222, Monday, December 3, 2012 at 12:46am
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.