Electrolysis of molten MgCl2 is the final production step in the isolation of magnesium from seawater by ....?

Electrolysis of molten MgCl2 is the final production step in the isolation of magnesium from seawater by the Dow process. When 34.4 g Mg metal forms, calculate the following.


(a) How many moles of electrons are required?





(b) How many coulombs are required?





(c) How many amps are required to produce this amount in 3.00 h?





please help.Electrolysis of molten MgCl2 is the final production step in the isolation of magnesium from seawater by ....?
Molar mass of Mg = 24.31 g/mol


Mole of Mg produced = 34.4 g / 24.31 g/mol = 1.415 mol





(a) Reduction equation;


Mg^2+(aq) + 2e- ------%26gt; Mg(s)





For each mole of Mg produced, 2 mole of e- are required.


For 1.415 mole of Mg; 2 x 1.415 = 2.83 mol e- are required.





(b) The charge of 1 mol e- is 1 F which is 96485 coulombs,


The charge of 2.83 mol e- ;


2.83 mol e- x 96485 coulombs / mol e- = 273,053 coulombs.





(c) Ampere = Coulombs / second





Time = 3.00 h x 3600 sec / hr = 10800 sec


Ampere = 273053 / 10800 = 25.28 A