Sciencemadness Discussion Board

Can Someone Help Me With Stoichiometry?

K12Chemistry - 16-2-2013 at 10:00

I am trying to make sodium iodide with crystal Iodine and sodium hydroxide. Can someone basically tell me how much iodine to how much sodium hydroxide and how much water I dissolve it in.

I am a noob :(

Stoichiometry

Bot0nist - 16-2-2013 at 10:56

No, but I will "teach you to fish" so to say.

http://cavemanchemistry.com/oldcave/projects/stoich/index.ht...

It's the "boring" bits of chemistry, but it is invaluable, and once you know, it's like riding a bike.

That whole site is great, and helped me learn chemistry many years ago. Work through all the lessons and projects, and you'll be surprised by how much you'll advance.

Good luck,
The Bot.

[Edited on 16-2-2013 by Bot0nist]

SM2 - 16-2-2013 at 15:09

moles and MW. equivalences. Don't forget Avogadros' #!

chemrox - 16-2-2013 at 15:23

I don't see how Avogadro's number comes into it but the mole concept is the core. Avogadro's just tells you how many molecule you have in a mole. It doesn't matter as long as you know how many moles you're dealing with. Look up and make sure you understand dimensional analysis. I'm old. I learned to do it on a slide rule. The beauty was in setting up the DA so you simply move the slide or the rule for each step and kept track of the 10's. This is the root of chemistry. Don't neglect it or all the rest wll baffle. I can't count the number of times I had a student tell me, "I love the chemistry but can't dig the math.." WTF??

Magpie - 16-2-2013 at 16:05

According to Wki, this is not all that simple, ie:

In basic solutions (such as aqueous sodium hydroxide), iodine converts in a two stage reaction to iodide and iodate:

I2 + 2 OH− → I− + IO− + H2O
3 IO− → 2 I− + IO−3


to simplify, multiply the first equation by 3 :

3I2 + 6OH- --> 3I- +3IO- +3H2O
3 IO- --> 2I- + IO3-

Then add the 2 equations giving:

3I2 + 6OH- --> 5I- + 3H2O + IO3-

For your use this would be:

3I2 + 6NaOH ---> 5NaI + 3H2O + NaIO3

Therefore 3 moles of I2 react with 6 moles of NaOH to give 5 moles of NaI, 1 mole of NaIO3, and 3 moles of water.

A g-mole is the weight of a molecule in grams divided by its molecular weight (MW). You can look up the MW's in a handbook or calculate them from the atomic weights shown in the periodic table.

The amount of water you use is up to you. How concentrated (or dilute) do you want your solution?





[Edited on 17-2-2013 by Magpie]

K12Chemistry - 17-2-2013 at 05:47

does the iodate precipitate out?

also how many grams of iodine to how many grams of NaOH?

Bot0nist - 17-2-2013 at 05:54

Please show an attempt to calculate the amounts yourself, and we will correct you if there is problems. People didn't post all this great learning material here, for you to just request the spoon.

Good luck.

blogfast25 - 17-2-2013 at 06:51

Quote: Originally posted by K12Chemistry  
does the iodate precipitate out?

also how many grams of iodine to how many grams of NaOH?


No, when you reduce the solution (by boiling in), the iodate and iodide both crystallise out. The iodate is then converted to iodide by heating: like chlorate it loses O2:

KIO3 === > KI + 3/2 O2

After that step you'll need to recrystallise your crude KI to get decent purity: saturate hot distilled (or deionised) water with the crud, filter if needed, allow to cool and chill to obtain a crop of clean KI crystals. Wash these with iced clean water and dry tham. Consult Wiki's solubility table to get an idea of quantities to dissolve.

Calculate the amount of I2 from Magpie's overall equation which shows you need 1 Relative Molar Mass of I2 for every 2 Relative Molar Masses of NaOH.



[Edited on 17-2-2013 by blogfast25]

K12Chemistry - 17-2-2013 at 11:13

umm, is it 2.5g of NaOH to 10g of iodine?

is this right?

Magpie - 17-2-2013 at 20:52

Quote: Originally posted by K12Chemistry  
umm, is it 2.5g of NaOH to 10g of iodine?

is this right?


Not quite. Show us your work and we will critique it. ;)

K12Chemistry - 18-2-2013 at 08:21

Hey guys,

I'm only 12 and you are helping me to learn chemistry a lot, thanks :)

OK so I looked and the caveman to chemistry link and tried it. It's probably wrong but here goes.

3I2 + 6NaOH ---> 5NaI + 3H2O + NaIO3

3 iodine moles = 380.71341g?
6 NaOH ~ 240g?

so umm,
simplified is it 19g of iodine to 12g of sodium hydroxide?

Sorry if it is wrong I'm learning...

[Edited on 18-2-2013 by K12Chemistry]

blogfast25 - 18-2-2013 at 09:44

Correct. Well done.

Now work out how much NaI you would get (theoretically) from those quantities.

[Edited on 18-2-2013 by blogfast25]

woelen - 18-2-2013 at 09:52

Better work out how much NaI you first get (only 5/6 of the iodine ends up in the iodide, the rest in the iodate). On heating, the NaIO3 looses oxygen and so you can also compute how much NaI you'll end up in the end.

In practice you need to recrystallize the solid after heating, because you also will loose some iodine (purple vapor) when you heat the NaI/NaIO3 mix. You'll end up with mostly NaI which contains some NaOH as well.

K12Chemistry - 18-2-2013 at 10:24

woohoo! :D

K12Chemistry - 18-2-2013 at 10:27

I don't want to use 19g of iodine and I don't have a balance that can go into decimals. Any ideas?

Magpie - 18-2-2013 at 10:38

Quote: Originally posted by K12Chemistry  

3I2 + 6NaOH ---> 5NaI + 3H2O + NaIO3

3 iodine moles = 380.71341g?
6 NaOH ~ 240g?


Uh, you can't ignore the "2" on I2.

3 iodine moles = 3(2)(126.9) = 761.4g

Otherwise, you have the right idea. ;)

[Edited on 18-2-2013 by Magpie]

[Edited on 18-2-2013 by Magpie]

K12Chemistry - 2-4-2013 at 03:36

sorry to bump but is the equation shown by magpie the same for Potassium Hydroxide instead of Sodium?

plante1999 - 2-4-2013 at 03:44

Short answer, yes.

K12Chemistry - 4-4-2013 at 10:52

long answer?

DraconicAcid - 4-4-2013 at 11:04

Quote: Originally posted by K12Chemistry  
long answer?


3I2 + 6NaOH ---> 5NaI + 3H2O + NaIO3


3I2 + 6KOH ---> 5KI + 3H2O + KIO3

The equation is the same (save for the replacement of one alkali metal with another). The molar masses used in the calculations will be different.

K12Chemistry - 5-4-2013 at 07:20

Does anyone know of a method to separate KI from KIO3? It doesn't precipitate like it would in water because it dissolves in KI solution. Heating strongly is not an option for me because I don't have the equipment.

Also how do you get the number to go down?

Like in DraconicAcids answer.

You know like instead of 3I2 it would be 3I(small 2)

DraconicAcid - 5-4-2013 at 07:54

Quote: Originally posted by K12Chemistry  


Also how do you get the number to go down?

Like in DraconicAcids answer.

You know like instead of 3I2 it would be 3I(small 2)


As you type, you'll notice the buttons above the typing window- B, I U, etc. On the right hand side, there is Xy and Xy. You can use those buttons.

Alternatively, you can type [s u b] before the numbers you want subscripted, and [/ sub] after them (no spaces- I just put the spaces in so that they don't actually turn into subscripts.

K12Chemistry - 5-4-2013 at 13:12

O2

DraconicAcid - 5-4-2013 at 13:14

Quote: Originally posted by K12Chemistry  
O2

Exactly!

Eddygp - 5-4-2013 at 13:24

Code:
[sub]...text...[/sub]

K12Chemistry - 6-4-2013 at 09:48

Now can someone tell me how to separate KI and KIO3.