I am planning to synthetisize clinoptilolite following the recipe of Gon Seo et al. (Catalysis Letters 55 (1998) 105–112). What amount of clinoptilolite can I obtain (if it works all fine)?
Ingredients:
3.22 g aluminum hydroxide hydrate (Al(OH)30.5H2O, Aldrich)
1.98 g NaOH (50% aqueous solution, Aldrich)
13.56 g KOH (45% aqueous solution, Aldrich)
50.0 g of aqueous colloidal silica (Ludox AS-40, DuPont)
32.35 g of deionized water
0.34 g seed crystals
Many thanks in advance!
Dear Kristina
I don't think it is easy to predict the amount of final product, unless someone who synthesized this product before helps you. I believe it is more suitable that you synthesize Clinoptilolite with the same ratio of reported literature as a model reaction, then if you need more Clinoptilolite you can easily prepare it (of course this time you'll be able to synthesize as much as you want).
Best of luck
Kindly go through the reference and follow the same proportions to get the product as obtained by them. I am surprised why do you want to synthesize clinoptilolite? It is freely available in many countries with different compositions. What is the purpose of this synthesis?
You have to find the product elemental composition. The ratio in percentage between reactant elemental composition and product elemental composition of the same elements give yield. Elemental composition of the product can be find out either as-synthesized or calcined. If you use calcined samples that will ideal result.
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