In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na,AlF), resulting in the reduction of the Al,0, to pure aluminum. Suppose a current of 6100. A is passed through a Hall-Heroult cell for 64.0 seconds. Calculate the mass of pure aluminum produced. Be sure your answer has a unit symbol and the correct number of significant digits.

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Chapter21: Chemistry Of The Main-group Elements
Section: Chapter Questions
Problem 21.163QP
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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (\( \text{Al}_2\text{O}_3 \)) dissolved in molten cryolite (\( \text{Na}_3\text{AlF}_6 \)), resulting in the reduction of the \( \text{Al}_2\text{O}_3 \) to pure aluminum.

Suppose a current of 6100. A is passed through a Hall-Heroult cell for 64.0 seconds. Calculate the mass of pure aluminum produced.

Be sure your answer has a unit symbol and the correct number of significant digits.
Transcribed Image Text:In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (\( \text{Al}_2\text{O}_3 \)) dissolved in molten cryolite (\( \text{Na}_3\text{AlF}_6 \)), resulting in the reduction of the \( \text{Al}_2\text{O}_3 \) to pure aluminum. Suppose a current of 6100. A is passed through a Hall-Heroult cell for 64.0 seconds. Calculate the mass of pure aluminum produced. Be sure your answer has a unit symbol and the correct number of significant digits.
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