How many coulombs of charge are required to produce 1 kg of Al metal from Al_2O_3?
The reaction required:
Al2O3 = 2Al3+ + 3O2-
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Al3+ + 3e- = Al(s)
1 kg of Al(s) = 1000 g of Al(s)
mol of Al(s) = mass/MW = 1000/26.981539 = 37.0623 mol of Al(s)
1 mol of Al = 3 mol of e-
37.0623 mol of Al = 3*37.0623 = 111.1869 mol of e-
recall that faraday calculated the charge of 1 mol of e-:
1 mol of e- = 96500 C
111.1869 mol of e- = 111.1869(96500) = 10,729,535.85 C required