How many coulombs of charge are necessary to produce 101 g of Al from Al³⁺(aq)? (F = 96,500 C/mol)

Question

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How many coulombs of charge are necessary to produce 101 g of Al from Al³⁺(aq)? (F = 96,500 C/mol)

Answer 1

Al³⁺(aq) + 3e- ==> Al(s)

atomic mass Al = 26.98 g / mol

mols Al = 101 g x 1 mol / 26.98 g = 3.744 mols

mols electrons (e-) = 3.744 mols Al x 3 mols e- / mol Al = 11.23 mols of electrons

Coulombs = 11.23 mols e- x 96,500 C / mol = 1,083,695 coulombs