LuckyWinner
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How much NaOh to use to make Sodium Iodide?
the formula for this reaction is:
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(6)NaOH + (3)I2 --> NaIO3 + (5)NaI + (3)H2O
239.955g + 761.42g --> 197.8924 g + 749.45g + 3 H2O
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Sodium hydroxide
39.992509 g/mol
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Iodine I2
Molecular Weight, 253.8089 g/mol
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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?
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MidLifeChemist
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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.
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Maurice VD 37
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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.
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MidLifeChemist
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Sometimes it is better to watch than to imagine.
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LuckyWinner
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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]
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