What amount of current, in amperes, would be required to deposit 19.7 g of Au metal in 1.00 hour from a solution of Au(NO3)3?

You can deposit 196.97/3 = approx 66 g Au with 96,485 coulombs of electricity. So how much will be required to deposit 19.7g

96,485 x (19.7/66) = approx 29,000 coulombs.
coulombs = amperes x seconds.
Substitute coulombs and seconds and solve for amperes.
Note those numbers I used are approximate. You need to redo all of them.

To determine the amount of current required, we can use Faraday's Law of Electrolysis. The formula is as follows:

1. Amount of Substance (in moles) = Mass of Substance (in grams) / Molar Mass of Substance

For gold (Au), the molar mass is 196.967 g/mol. Let's calculate the number of moles of Au:

Amount of Substance (in moles) = 19.7 g / 196.967 g/mol

Next, we need to use Faraday's Law to relate the amount of substance to the amount of electricity passed through the solution:

2. Amount of Electricity (in coulombs) = Amount of Substance (in moles) x Faraday's Constant

The Faraday's constant is approximately 96485 C/mol.

Let's calculate the amount of electricity required:

Amount of Electricity (in coulombs) = (19.7 g / 196.967 g/mol) x 96485 C/mol

Now, since we know the time in hours, we need to convert it to seconds:

1 hour = 3600 seconds

Finally, we can calculate the current in amperes using the formula:

3. Current (in amperes) = Amount of Electricity (in coulombs) / Time (in seconds)

Let's substitute the values into the equation to find the current:

Current (in amperes) = [(19.7 g / 196.967 g/mol) x 96485 C/mol] / (1 hour x 3600 seconds/hour)

By evaluating this equation, we can determine the current required to deposit 19.7 g of Au metal in 1.00 hour from a solution of Au(NO3)3.