How many minutes will it take to electroplate 49.1 g of gold by running 5.00 A of current through a solution of Au+(aq)?

Au⁺(aq) + e⁻ → Au(s)
No. of moles of Au = (49.1 g) / (197 g/mol) = 0.24924 mol
No. of moles of e⁻ passed = 0.24924 mol

Each mole of e⁻ carries 96485 C of electricity.
Amount of electricity passed = (0.24924 mol) × (96485 C/mol) = 24048 C
Time taken = (24048 C) / (5.00 C/s) = 4810 s = 80.2min

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OR:
(49.1 g Au) × (1 mol Au / 197 g Au) × (1 mol e⁻ / 1 mol Au) × (96485 C / 1 mol e⁻) × (1 s / 5.00 C) × (1 min / 60 s)
= 80.2 min

Au is 197 g/mol
49.1 g / 197 g/mol = 0.249 mole

6.022e23 atoms/mole     Avogadro constant
6.022e23 atoms/mole x 0.249 mole = 1.501e23 atoms

each atom requires 1 electron to be plated, so that is 1.501e23 electrons.
1 Coulomb = 6.242e18 electrons
1.501e23 electrons x 1 C /  6.242e18 electrons = 24050 C

5 amps = 5 C/s
24050 C / 5 C/s = 4810 sec or 80.2 min