A copper electrode weighs 35.42 g before the electrolysis of a CuSO4(aq) solution and weighs

Question

36.69 g after the electrolysis has run for 20.0 min. What was the amperage (Amps) of the current
used?

this is the equation that i used is it correct

(1.27g Cu x 1 mol Cu x 2 mol e- x 96,485 c) / (63.55g Cu x 1 mol Cu x 1 mol e- x 20 min x 60 sec)

= 0.002528 A

Answer 1

I didn't check out our equation but I don't get that answer. INstread I plugged your answer in to see if it would plate out 1.27 grams, as follows:
Coulombs = 0.002528 x 20 x 60 = approx 3 coulombs.
96,485 coulombs will plat out 63.55/2 = 31.775 g Cu so
31.775 x about 3.0/96,485 = about 1E-3g or approx 1 mg but the problem plates out 1.27 g so 0.002528 A is not right. The correct answer is closer to 3 amperes.