Chemistry - Synthesis of Potassium Trioxalatoaluminate (III) Trihydrate

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 ;)

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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.

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