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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: albert611 on August 29, 2004, 04:35:02 PM

Title: Formation of Alum [KAl(SO4)2·12H2O]
Post by: albert611 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.

Title: Re:Formation of Alum [KAl(SO4)2·12H2O]
Post by: AWK 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.