Al₂O₃ + 6 e⁻ + 6 H⁺ → 2 Al(s) + 3 H₂O
Q = i * t = (1.10 x 10⁶ Amp)*(2.80 x 60 x 60) = 1.11 x 10¹⁰ C
number of moles = 1.11 x 10¹⁰ C / 96485 C / mol = 1.15 x 10⁵ mole of e⁻
6 : 2 (ratio of e and Al)
1.15 x 10⁵ : ??
mole of Al = 1.15 x 10⁵ x

= 3.83 x 10⁴ mole of Al
Molar mass of Al = 27 g / mole
Mass of Al produced = 3.83 x 10⁴ mol x 27 g/mole = 1.03 x 10⁶ g