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We mixed $\ce{Al}$ powder and hot water. Then we added $\ce{KOH}$ to the $\ce{Al}$ solution and it bubbled heavily (producing $\ce{H2}$). 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:

$$\ce{Al + KOH + 2H2O -> K[Al(OH)3] + H2}$$

and when oxalic acid comes in:

$$\ce{K[Al(OH)3] + C2H2O4.2H2O ->K3[Al(COO)3].3H2O}$$

The problem is this site says that the reaction is wrong and What is the reaction between oxalic acid and potassium permanganate? is a similar question but has a completely different looking answer. I'm lost.

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  • $\begingroup$ Welcome to Chemistry.SE. I like that you described the steps and your observations! You're not lost, only the stoichiometry is a bit off. That can be fixed :) $\endgroup$ – Klaus-Dieter Warzecha Feb 20 '16 at 7:25
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You have already concluded that the bubbles indicate that hydrogen gas is formed. Let's assume that two protons are reduced. Each receives an electron and they form a molecule of $\ce{H2}$:

$$\ce{2H+ + 2e- -> H2}$$

Where did the electrons come from? Did the aluminium powder dissolve?

$$\ce{Al -> Al^3+ + 3e-}$$

We are now having two equations, one for a reduction, one for an oxidation (of aluminium).

Let's add these two equations in a way that the electrons cancel out. For that, we multiply the first equation by 3, and the second equation by 2.

$$\ce{2Al +6H+-> 2Al^3+ + 3H2 ^}$$

Something's still wrong with the equation: You didn't throw the aluminium powder into a sea of protons, but suspended it in hot water. Let's add some $\ce{OH-}$ on both sides:

$$\ce{2Al +6H2O -> 2Al(OH)3 + 3H2 ^}$$

This is much better, but still not perfect: You did not describe a cloudy precipitate of a hydroxide being formed, and your first step took place in the presence of potassium hydroxide:

$$\ce{2Al +6H2O +2KOH-> 2K[Al(OH)4] + 3H2 ^}$$

So far, we have explained the reaction of aluminium metal in aqueous alkaline solution: tetrahydroxoaluminate $\ce{[Al(OH)4]-}$ is formed.

Can you figure out the reaction with the oxalic acid on your own? Have a look at the structure of oxalate (Hint: bidentate ligand) and forget about the permanganate ;)


UPDATE

Let's add some oxalic acid, $\ce{HOOC-COOH}$ now. For the reaction, it doesn't make any difference whether it's a dihydrate, $\ce{H2C2O4*2H2O}$, or not. The reaction is performed in aqueous solution anyway. However, you might want to consider the difference in molecular weight between the anhydrous form and the dihydrate when you're planning to add a stoichiometric amount of oxalic acid.

$$\ce{[Al(OH)4]- + 3HOOC-COOH -> [Al(OOC-COO)3]^3- +2 H+ + 4H2O}$$

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  • $\begingroup$ So I have three oxalates that are bound to Al, each in a ring form, and 3 K's to make the total charge 0? What happens to the 4 OH's and the 6 H+'s that were taken off oxalic acid to make it into oxalate? I feel like putting them together to form water but I'm not sure since it's not a perfect match. How does the fact that oxalic acid is a dihydrate play into all this? I'm sorry but I need more guidance. $\endgroup$ – Kra Feb 21 '16 at 0:22
  • $\begingroup$ This is perfect! Thank you so much. It looks like I was overcomplicating it by not combining my OHs and H, and the part about dihydrate makes a lot of sense. Thanks again for the detailed response! $\endgroup$ – Kra Feb 22 '16 at 2:59
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I think the reaction takes place in such a way that:

$\ce{Al + 3KOH -> K3AlO3 + 3/2 H2}$

And then after addition of Oxalic acid:

$\ce{K3AlO3 + 3H2C2O4 -> K3[Al(C2O4)3]H2O + 2H2O}$

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