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Topic: Formation of Alum [KAl(SO4)2·12H2O]  (Read 12024 times)

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albert611

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Formation of Alum [KAl(SO4)2·12H2O]
« on: August 29, 2004, 04:35:02 PM »
Just did a lab about synthesizing alum [KAl(SO4)2·12H2O], but I don't understand how to write a balanced chemical equation for it. Heres a brief rundown of the procedure:

1. Add 25 mL 3M KOH to aluminum foil.
2. Acidify with 35 mL 3 M H2SO4.
3. Filter solution and boil.
4. Cool and allow crystals to form.

I tried balancing it with the equation that I got, but it didn't work. Thanks.

Offline AWK

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Re:Formation of Alum [KAl(SO4)2·12H2O]
« Reply #1 on: August 30, 2004, 02:06:37 AM »
2Al + 2KOH + 6H2O = 2K[Al(OH)4 ] + 3H2(g)

K[Al(OH)4 ]  + 2H2SO4  + 2H2O= K[Al(H2O)6 ](SO4)2

Note,  in solution and in solid Al cation is hydrated, in the solid the rest of  H2O molecules  constitutes a crystallization water.
« Last Edit: August 30, 2004, 02:15:30 AM by AWK »
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