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Question
How much electricity in terms of Faraday is required to produce 40.0 g of \[\ce{Al}\] from molten \[\ce{Al2O3}\]?
(Given: Molar mass of Aluminium is 27 g mol−1.)
Solution
\[\ce{Al2O3 -> 2Al^{3+} + 3O^{2-}}\]
27 g of aluminium needs = 3 mole of electrons
= 3 × 96500 coulombs
∴ 40.0 g of aluminium needs = `(3 xx 96500 xx 40.0)/27`
= 4.28888 × 105 coulombs (4.44 F)
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