[Mustungun]

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

[chenbeier]

I would eleminate some water in your equations. Zirconiumhydroxide will not carry 6 water.

[Mustungun]

You're right. Does the rest seem to be correct?

[chenbeier]

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.

[Mustungun]

ZrO(NO₃)₂*6H₂O + 2NaOH ---> Zr(OH)₄ + 5H₂O + 2NaNO₃

How's this?

[chenbeier]

Sound good.

[Mustungun]

Perfect. Thank you alot.