Chem
posted by Ginger .
How many minutes will it take a current of 66.049 Ampères to deposit 86.825 grams of aluminum (26.981538 g/mol)?
Al has a charge of 3, so n=3.
MM=26.981538g/mol
1eq=molarmass/charge
so 26.981538/3=8.9938
1eq=8.9938
1F=96485c/eq
F=13138.29477
Am I on the right path? I don't know where to go from here...

26.98 g Al/3 = 8.994 g = mass deposited by 96,485 coulombs.
To deposit 88.825 g, we must have
?C = 96,485 x 88.825/8.994 = ??
Amperes x seconds = Coulombs.
You have C and you have A, solve for seconds and convert to minutes.