Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Mustungun on November 14, 2018, 09:20:47 AM
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Hi everyone. I'm doing an internship at my university in the industrial chemistry branch. I'm currently working on a hard template synthesis of zirconium hydroxide.
I know the precursor is ZrO(NO3)2 * 6H2O and it somehow becomes Zr(OH)4, which is needed for the sulphation process. Though, the calcination needed to remove the template transforms it into ZrO2, which is kinda problematic for the sulphation, but I digress. I'm trying to figure out how should I write the reactions. I have an hypothesis, but I can't confirm it for myself. Can you confirm it for me, or correct me if I'm wrong? Thank you in advance for your time.
2ZrO(NO3)2* 6H2O + 4NaOH + 4H2O + O2 ---> 2Zr(OH)4 * 6H2O + 4NaNO3
Zr(OH)4*6H2O ---> ZrO2 + 8H2O
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I would eleminate some water in your equations. Zirconiumhydroxide will not carry 6 water.
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You're right. Does the rest seem to be correct?
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The equations i still not ok, because there is no redox reaction. Zirconium stays with oxidation number 4. So you dont need oxygen as well.
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ZrO(NO3)2*6H2O + 2NaOH ---> Zr(OH)4 + 5H2O + 2NaNO3
How's this?
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Sound good.
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Perfect. Thank you alot.