How many coulombs of charge are necessary to produce 345 g of Al from Al³⁺(aq)? (F = 96,500 C/mol)

Answer 1

Al3+(aq) + 3e- ==> Al(s)

345 g Al x 1 mol Al / 26.98 g = 12.79 mols Al

12.79 mols Al x 3 mols e- / mol Al = 38.36 mols electrons needed

38.36 mols e- x 96,500 C / mol e- = 3,702,000 C needed = 3.70x10⁶ C (3 sig. figs.)