guy - 20-5-2005 at 22:34
Which is the correct rxn for Al with NaOH?
2NaOH + 2Al + 6H20 --> 2NaAl(OH)4 + 3H2
or
NaOH + Al + 2H2O --> NaAlO2 + 2H2
I found the first equation in my textbook.
I dissolved Al foil (5g) in NaOH (7g) and filtered off a lot of grayish powder(?impurities). In the end I obtained a light yellow liquid. Which is
the product?
[Edited on 5/21/2005 by guy]
sparkgap - 20-5-2005 at 22:48
Something seems missing... isn't H<sub>2</sub> gas supposed to be evolved in this reaction?
But yes, the Al becomes Al(OH)<sub>4</sub><sup>-</sup> upon completion.
sparky (~_~)
guy - 20-5-2005 at 22:56
How come so many websites say that sodium aluminate is NaAlO2, or is NaAl(OH)4 just the hydrated form?
sparkgap - 20-5-2005 at 23:01
It's actually more of a complexation than a hydration in the particular reaction you mentioned, because it's in solution. The
OH<sup>-</sup> ions donate their electrons to Al's empty d orbitals, IIRC.
I don't think aluminate is the right way of naming that anion; it's more like tetrahydroxyaluminate(III). (gee, was that pedantic? 
)
sparky (^_^)
12AX7 - 21-5-2005 at 01:09
The extra OH is just water, NaAlO2 (anh.) + 2H2O = NaAl(OH)4. The latter form is more correct since aluminum complexes with water (AlCl3 cannot be
dehydrated!), but can often be ignored in reactions since the water just goes along for the ride.
Tim
chemoleo - 21-5-2005 at 10:53
For the record, Na-aluminate is also formed by the direct reaction between NaOH (s) and Al.
This has been discussed (extensively) in the 'unconventional sodium' thread, as a means of making Na, that supposedly forms in due course,
too.