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3 years ago

The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3 * xH2O.

Chemistry

1 answer:

lesantik [10]3 years ago

6 0

Answer:

44.7 kWh

Explanation:

Let's consider the reduction of Al₂O₃ to Al in the Bayer process.

6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻

We can establish the following relations:

The molar mass of Al is 26.98 g/mol.
2 moles of Al are produced when 6 moles of e⁻ circulate.
1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
1 V = 1 J/c
1 kWh = 3.6 × 10⁶ J

When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

$3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh$

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