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Question

How many coulombs are required to deposit 5.4 g of aluminum from an AlCl₃ solution? (Atomic mass of Al = 27 g/mol, F = 96500 C/mol)

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Explanation

Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (5.4/27) = 0.2 mol , Charge = 0.2 × 3 × 96500 = 57900 C .

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