Homework Help: Chem

Posted by Ginger on Sunday, March 30, 2008 at 6:46pm.

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...

Chem - DrBob222, Sunday, March 30, 2008 at 7:47pm
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.

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