How many amperes of current are required to deposit 5.00 g of gold on the cathode in 60.0 minutes from a solution containing a salt of gold (III)?

=      2.04 amps

Explanation:

5.00 g Au = 5.00 g Au X (1.00 mol Au/197 g Au) X (3 mole e⁻/mole Au) X (96,500amp·sec/mole e⁻) X (1min/60sec)  X (1/60min)                                

=      (5.00)(1.00/197)(3.00/1)(96,500/1)(1/60)(1/60)amps

=      2.04 amps

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