Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: paragoon on February 19, 2016, 11:03:42 PM
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We mixed Al powder and hot water. Then we added KOH to the Al solution and it bubbled heavily (producing H2). When that was over we added oxalic acid dihydrate to the boiling aluminate solution. We filtered and added cold ethanol. How do write these reactions?
I was thinking Al + KOH + 2H2O = K(Al(OH)3) + H2
and when oxalic acid comes in
K(Al(OH)3) + C2H2O4*2H2O = K3(Al(COO)3)*3H2O
The problem is, http://www.webqc.org/balance.php says that it's wrong and http://chemistry.stackexchange.com/questions/4310/what-is-the-reaction-between-oxalic-acid-and-potassium-permanganate?lq=1? which is kind of similar has a completely different looking answer. I'm lost.
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Good starting point would be to write correct formulas for every compound involved.
For example, no such thing as K(Al(OH)3).
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Good starting point would be to write correct formulas for every compound involved.
For example, no such thing as K(Al(OH)3).
How about this? 2Al + 2KOH + 6H2O = 2K(Al(OH)4) + 3H2
Now I want 2K(Al(OH)4) + C2H2O4 ∙ 2H2O to make K3Al(C2O4)3 ∙ 3H2O but I'm unsure on how to proceed. I thought of doing a double replacement but that's clearly wrong. Judging by the product it appears that three oxalates bind to Aluminum and leave the Potassium untouched but I don't know why. How are three oxalates bound to Aluminum when each one has two O-?
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Potassium is not bound to anything. Product is a dissolved salt, fully dissociated.
Do you know what is a net ionic reaction?
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Potassium is not bound to anything. Product is a dissolved salt, fully dissociated.
Do you know what is a net ionic reaction?
Yes, I have done those so Potassium is a spectator that doesn't participate in the intermediate reactions?
Just to add to my last post:
I just learned that the cause is something called the Chelate Effect. Two O-'s from each of the three oxalates binds to Aluminum. The Al core has 3+ and the O-'s are 6- (since there are six of them). That is why we have K3 because K+ times three = 3+ to make the final charge 0. But what about the 4 OH-'s from the beginning? I was thinking that the H+'s the were removed from oxalic acid to make oxalate would join with the OH-'s to make water but not sure.
I have K3Al(C2O4)3 ∙ 3H2O + 4OH- + 6H+ as final products but something seems wrong.
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so Potassium is a spectator that doesn't participate in the intermediate reactions?
Yes.
4OH- + 6H+
Can they exist in a solution at the same time?
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so Potassium is a spectator that doesn't participate in the intermediate reactions?
Yes.
4OH- + 6H+
Can they exist in a solution at the same time?
I see so it is 2H+ and 4H2O. I understand it now. Thank you.