In the hall-heroult process, a large electric current is passed through a solution of aluminum oxide (a12o 3) dissolved in molten cryolite (na3aif6) ,resulting in the reduction of the ai�3 to pure aluminum. suppose a current of 6800.a is passed through a hall-heroult cell for 54.0 seconds. calculate the mass of pure aluminum produced. be sure your answer has a unit symbol and the correct number of significant digits.

34.2 g

Explanation:

In the Hall-Heroult process, Al³⁺ (from Al₂O₃) is reduced to Al. The reduction half-reaction is:

Al³⁺ + 3 e⁻ ⇒ Al

We can establish the following relations:

1 A = 1 c/s1 mole of e⁻ has a charge of 96468 c (Faraday's constant)1 mol of Al is produced when 3 moles of e⁻ circulateThe molar mass of Al is 26.98 g/mol.

Suppose a current of 6800 A is passed through a Hall-Heroult cell for 54.0 seconds. The mass of Al produced is:

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