Sciencemadness Discussion Board

How much NaOh to use to make Sodium Iodide?

LuckyWinner - 19-11-2020 at 12:31

the formula for this reaction is:
--------------------------------------------
(6)NaOH + (3)I2 --> NaIO3 + (5)NaI + (3)H2O
239.955g + 761.42g --> 197.8924 g + 749.45g + 3 H2O
--------------------------------------------
Sodium hydroxide
39.992509 g/mol
--------------------------------------------
Iodine I2
Molecular Weight, 253.8089 g/mol
--------------------------------------------

Question:

to get to 10g of Iodine used we
divide everything by 76.1 =
-10g Iodine
-3.15g of NaOh

now why does every youtube video say you should use 2.5g?
thats 21% less NaOh.

https://youtu.be/7vZOJRPnY8M
https://www.youtube.com/watch?v=OkUszT4tFww

both use 2.5g of NaOh for 10g Iodine


what is the right amount to use?
and why does using sodium carbonate not work?

MidLifeChemist - 20-11-2020 at 09:14

Many of the youtubers just copy what other people have posted. So if one person posts something that is incorrect, the others copy the same mistake. I haven't done this particular experiment, but personally I would go with your own math.

Maurice VD 37 - 21-11-2020 at 12:11

With your calculation, you are losing 16.6% of the Iodine involved. because for every 5 NaI produced, you also produce 1 NaIO3. And the iodine fixed in this NaIO3 is lost. I have not seen the other videos published on Youtube, but I imagine that they add something to reduce NaIO3 into NaI. If you are able to do this with an unknown product X, the equation becomes . 2 NaOH + I2 + X --> 2 NaI + ... XO2H2(?). And with this equation, 10 g I2 is reacted with 2.5 g NaOH.

MidLifeChemist - 21-11-2020 at 13:52

Quote: Originally posted by Maurice VD 37  
I have not seen the other videos published on Youtube, but I imagine that


Sometimes it is better to watch than to imagine.

LuckyWinner - 21-11-2020 at 14:02

Quote: Originally posted by Maurice VD 37  
With your calculation, you are losing 16.6% of the Iodine involved. because for every 5 NaI produced, you also produce 1 NaIO3. And the iodine fixed in this NaIO3 is lost. I have not seen the other videos published on Youtube, but I imagine that they add something to reduce NaIO3 into NaI. If you are able to do this with an unknown product X, the equation becomes . 2 NaOH + I2 + X --> 2 NaI + ... XO2H2(?). And with this equation, 10 g I2 is reacted with 2.5 g NaOH.


apparently NaIO3 decomposes to NaI at ~425C.
just put it in a furnace at 450C for a while.
NaIO3's Melting point: 425 °C

otherwise oxidize it back to elemental iodine,
or does that not work with NaIO3 ?


and the youtubers do not add anything to reduce it,

if you add an oxidizer to the NaI ,NaIO3 mix , you ll also reduce the NaI back to elemental iodine, no?

I tried it with my 1 to 1 mol calculation already and it seemed to
give me a perfect yield.



[Edited on 21-11-2020 by LuckyWinner]

[Edited on 21-11-2020 by LuckyWinner]