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Need help understanding KOH + Al2O3 reaction

  1. May 8, 2009 #1
    Sorry if this is a dumb question but I'm an ME and am only now starting to get a grip on chemistry.

    If I have ruby (Cr:Al2O3) in a humidified air environment that contains KOH, my limited chemistry knowledge tells me that the alkali, KOH, will react with the ruby. However, I can't figure out what the products of this reaction are. Since there is no hydrogen in the alumina, it can't be a acid-base reaction. So, I'm going to guess its a double placement reaction. Based on this, I get the following unbalanced reaction;

    KOH + Cr:Al2O3 -> K2O + Al(OH)3 + Cr(OH)3

    However, this just doesn't seem right to me as K2O is a rather unstable compound. Am I doing this right? Any help would be much appreciated.
     
  2. jcsd
  3. May 8, 2009 #2
    Actually, it is just an acid-base reaction. The acid is Al2O3. Alumina is an "amphiprotic" compound, which means that it can act as base, with strong acids, and as acid, with strong bases. For example, you can also dissolve alumina (always in the form of powder, or it's a very slow process) with K2S2O7, which releases SO3 (a very acid compound) at high temperatures (around 300°C) forming aluminum sulphate.

    Your reaction is:

    2KOH + Al2O3 + 3H2O --> 2KAl(OH)4

    Cr can substitute Al in that reaction.
     
    Last edited: May 8, 2009
  4. May 11, 2009 #3
    Thanks for the reply LightArrow, it would have taken me forever to figure that out. So, I'm assuming that if there is no water in the mix, a reaction would never take place?

    On another note, why did this thread get moved? Its not a course work question.
     
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