What mass of Al can be deposited by the passage of a constant current of 5.00 A through a Al(NO3)2 solution of 2.00 hours?
My work:
Q= I*t
= 5 * 7200sec = 36000
36000/96458= .3732
.3732* 1/3 * (26.98*3) = 10.06g
But this is wrong it's suppose to be 3.3.6g. why is that?
3 answers
where did the three in the last parenthesis come from.
I'm sorry I made a typo it goes through a solution of Al(NO3)3 and that's why I mutliplied the m.w. of Al by 3 because you have 3 moles of Al per NO3
It isn't the nitrate that is plating out. It's the Al. Remove the 3. You have taken care of the 3e in the transfer of Al(III) to Al(s) by the 1/3.