Posted by **Jennifer** on Saturday, March 31, 2007 at 8:44am.

Hw many minutes will it take a current of 31.559 Amperes to deposit 1.722 grams of aluminum?

I believe it to be 6.505 minutes

Is this correct? If not, I'll post my work. Thanks

I found 9.757 minutes.

Ok, here's my work, I just plugged in numbers to another problem, I hope this makes sense:

Convert grams to moles:

1.722g x 1g/26.981538mol =.0638214174mol

.0638214174mol x 2 = .1276428349 mol e-

.1276428349 mol e- x 965000 = 12317.53C

t = 12317.53C/ 31.559

t = 6.505

If you don't mind, can you explain where I went wrong. thanks

I tried copying this so I could mark the lines for errors but for some reason I can't copy it. I'll just explain it and you follow along.

#1. You used 2 for the number of electrons. Al is trivalent and the reaction is Al

^{3+} + 3e = Al. That is a major source of your error.

#2. I used 96485 for coulombs in an ampere instead of 96,500. If you are carrying so many decimal places as you show, then you should not round to 96,500; however, this is a minor source of error.

#3. You don't show the conversion to minutes, but I see your work reflects that.

So your 0.12764 will be 0.19146

and that times 96485 = 18,473

and that divided by 31.559 = 585.34 sec and that divided by 60 = 9.76 min.

Check my arithmetic

I didn't obey all the rules for significant figures. You should go through and round the final answer to the right number of s.f. Since 1.722 g has 4 s.f. I think you are allowed 4 in the answer if you use 96,485 and not 96,500.

Also, I forgot to mention that you have one too many zeros on 96500 but that is just a typo since the number doesn't reflect that extra zero.

Yeah I had a lot of errors. Thanks for explaing that.

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