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Yttrium2
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[*] posted on 13-3-2018 at 18:16
How much solution is required to react


This is a very basic question, it's been a while since my last chem class. (I've gotten rid of the book for some reason)

How do you set this problem up?


I went from 2g KI... x 1mole KI/166.0028g KI

This gave me the number of KI moles reacting, from there I went to see how many moles of H2O2 would be needed.

2g KI x 1moleKI/166.0028gKI x 1moleH2O2/2moleKI = the amount of moles of H2O2 needed to react with 2grams of KI, (.02409moles H2O2 needed)
From there I went to see how many milliliters of %3 solution is equivalent to .02409 moles of H2O2

So I did .02409 moles H2O2 x 34.047g H2O2/1 mole H2O2 x 100ml solution / 3 grams H2O2



Is this correct?
I found out how many moles H2O2 needed, then grams, then to milliliters of solution to get 27.3ml %3 H2O2 solution to react with 2 grams KI.






Did I do this correctly? I have a feeling I messed up where I went from grams H2O2 needed to the amount of milliliters of %3 needed.



Is this correct?
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[*] posted on 13-3-2018 at 18:35


Stoichiometry

The trick with these I tell my students is to make good headings. This helps you to keep track of things and also enables you to locate errors easily.

Balanced Equations
2I- = I2 + 2e-
H2O2 + 2H+ + 2e- = 2H2O

Combining and adding spectator ions
2KI + H2SO4 + H2O2 = K2SO4 + 2H2O

Moles of KI available
n=2/166 = 0.012

Moles of H2O2 needed
n= 0.012 ÷ 1 × 2 = 0.006
(use the coefficients from your balanced equation)

Mass of H2O2 needed
m=0.006 × 34 = 0.205

Mass of 3% solution needed
m=0.205 ÷ 0.03 = 6.83 grams or approx 6.8mL.
I'd add a maybe 10-20% excess.

In your calculation you have a four times excess resulting from multiplying by 2 instead of dividing.






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[*] posted on 21-4-2018 at 22:43


I'm had a hard time seeing understanding what you wrote. Particularly where you are getting your values from, and what the values are. I like to keep things labeled/use dimensional analysis...

So for 1.2g of Iodine, or potassium iodide....



...

1.2g I2 x 1 mole I2 per 253.8089g I2 gives moles I

this then is multiplied with the conversion factor of 1 over 1, so it stays the same

Therefore,
1.2g I2 needs .0047279665 moles H2O2

.0047 moles H2O2 = .159868996g H2O2


Now from there I set up a proportion to find out how many mL's of 3% H2O2 are needed to have .159868996g H2O2

3g H2O2 per 100ml X .159808996g per 'y' mL's

I got 3 y per 3 X 15.9808996g per 3

Final answer 5.33 mL's of 3% H2O2 required to react 1.2g I2.

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[*] posted on 21-4-2018 at 23:43


8 significant digits, wow! I usually think just in moles:
3 percent H2O2 is 0.88 Molar.
2 grams of KI is 12 mmol.
They react 2:1 so one needs 6 mmol H2O2
6 mmol/0.88 molar=6.8 mL
And I agree with jsum, add a small excess of oxidant. (which by the way means that you might as well do this calculation with only integer values). Dimensional analysis is great for working out what unit the value is in and as such comes secondary to proper headings.
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[*] posted on 1-5-2018 at 19:03


Okay, so apparently I'm still a little confused here. There are videos all over the net

showing how to get crystal iodine, but nothing I could find showing the actual

calculations behind it.

starting with oh, say 15g KI, how much iodine will this produce and how much 3%

hydrogen peroxide will it require

Additionally, how are they able to perform the measurements using volumetric

parts instead of gram to gram conversions?


I'm guessing the first step is to

A) convert grams of KI available to moles of KI available

B) use conversion factor to find out how many moles of other reacting reagents is

required / find out how many grams you need based off of the molar conversion

factor

C) Find out how many mL's of 3% h2o2 are required for the necessary amount of

h2o2 needed...





These are what my calculations look like, I'm stuck on number C, and also, I don't

understand how the problem can be done using seemingly simplistic volumetric

comparison.. (If that makes sense)




Can someone right it out for me as I'm having trouble uploading my calculations



Thanks

[Edited on 5/2/2018 by Yttrium2]
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[*] posted on 1-5-2018 at 20:00


Oh, you want elemental iodine? H2O2 doesn't work all that well for that, because KI catalyzes the decomposition of H2O2 into H2O and O2. And all the bubbles make it so that it's virtually impossible for you to ever collect the precipitate. What works better is to add chlorine bleach dropwise to a rapidly-stirred solution of KI and HCl. Let it settle occasionally, and if the solution has a greenish-yellow tint, that means all the iodine has come out of solution.

I think that the reason the videos don't explain the stoichiometry is because you're never sure of the concentration of the peroxide, and because a lot of it is lost due to the KI catalyzing its decomposition. Since those values are relative unknowns, any numbers you'd calculate aren't very meaningful anyway.




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[*] posted on 1-5-2018 at 20:03


What I'm trying to understand is the calculation process. We start with grams KI, then go to moles KI, and then we go to moles H2O2/moles KI, this gives us the amount of moles of H2O2 needed. From there we can go from moles needed to grams needed. Problem is, that it's a 3% solution. Does this mean simply multiplying the amount of H2O2 in grams needed by 3/100?




[Edited on 5/2/2018 by Yttrium2]
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[*] posted on 1-5-2018 at 23:35


Yep, solutions like this are usually either weight/weight or weight/volume. To figure out which it is I usually just google (if it is not defined on the bottle).

I got to it like this: I googled "weight/volume weight/weight peroxide", and got this hit: https://onlineconversion.vbulletin.net/forum/main-forums/con...
By this I figured 3% H2O2 is probably weight/weight.

Because it is weight/weight you either have to weigh the peroxide solution, or find the density to be able to measure the volume. The density is 1.0095 g/ml (15oC) (googled it). This is less than 1% percent off from 1 g/ml, so in this case you could use weight/weight and weight/volume interchangeable.

So yes, if you need e.g. 1.2 gram peroxide, and you have a 3% solution, divide 1.2 by 3 and multiply by 100. This gives you the weight of solution (40 grams) you need as 3% is 3 grams per 100 grams (weight/weight).

If you want the volume you can divide again by the density (1.0095), which gives you 39.62 ml.

Edit: If the 3% meant weight/volume, and you need 1.2 grams, you would have divided 1.2 by 3 and multiply by 100 as with the weight/weight, but it would have given 40 ml instead of 40 grams.

So 3% weight/weight is a little more concentrated than 3% weight/volume (in the case of hydrogen peroxide).




[Edited on 2-5-2018 by Tsjerk]
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[*] posted on 2-5-2018 at 02:28




[Edited on 2-5-2018 by Tsjerk]
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[*] posted on 2-5-2018 at 08:20


Good information, not so sure that it answers my questions however. I think it raises more questions than it answers, for me.

Could someone right out the dimensional analysis behind the problem and upload it, as I am having trouble uploading a pic of my dimensional analysis calculations.
I am having a hard time remembering how to do dimensional analysis.
I am having a hard time doing the problem.
I wonder if its even worth if for you to right it out for me if I can't even remember.



You said it would either have to be weighed or find the density to measure volume,

Why is this?

( I used to understand why this was, but somewhere over the last few years I seem to have forgotten)

How come one couldn't measure the volume of a 3% solution with a volumetric measuring device? Its 3 grams solute in 100mL of solution, right?

Is it because when you had 3 grams of solute into 100mL of solution the volume changes to be more than 100mL of solution?
so that this means to use a 100ml volumetric flask?
how again is a 3% weight weight solution made up?

I think we might be over complicating things a bit. but in either scenario, I'd like to have my original question answered. And now the additional questions answered from the added complexity.


Thank you for your time and consideration, (I usually am a bit unmindful when I say this, but when I say it now, I truly am grateful)

[Edited on 5/2/2018 by Yttrium2]

[Edited on 5/2/2018 by Yttrium2]
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[*] posted on 2-5-2018 at 08:41


If it's weight for weight, it's 3 grams in 97 grams of water.

How you gonna measure 100 grams with a volumetric flask without knowing the density?


Maybe you ought to try Melgar's suggestion. It doesn't require as much calculating, which you don't seem to be very into.

Also, the endpoint is indicated by the color of the solution.

If I ever need to convert KI to elemental I, I(meaning me) will probably try bleach first.

EDIT: In RE below post.
Then might I suggest a strengthening cocktail before work?
A dry martini with a bit of absinthe?

Perhaps grappa on ice?

Vodka and beef bullion?



[Edited on 2-5-2018 by SWIM]




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(Not as cool as it sounds)




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[*] posted on 2-5-2018 at 08:44


I'm getting weak, If I cant do it somebody else is going to have to do it for me.
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[*] posted on 2-5-2018 at 08:53


The density of 3% hydrogen peroxide is close enough to the density of water that it is common to ignore the difference in density and assume that the specific gravity is 1 unless doing something that demands extraordinarily high precision (and how precisely do you know the concentration of your hydrogen peroxide, really?).



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[*] posted on 15-5-2018 at 15:30


Bump can someone explain this problem?

I did an extraction, I'm not too impressed with my yield on first glance. (I haven't weighed it yet)


I'm wondering why my yield was so poor

The recipe I ended up using called for a minimum amount of water required to dissolve the KI, as much hcl as there is solution, and then seven times that amount of solution of h2o2

I eyeballed the amounts of solution, sometimes even using oval shaped glasses to try to do precise volumetric measurements in.


Out of 25grams KI, I only yielded what looks like to be a few grams of I2.


I didn't really wait for it to precipitate that long before filtering.



[Edited on 5/15/2018 by Yttrium2]

[Edited on 5/15/2018 by Yttrium2]
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[*] posted on 15-5-2018 at 16:49


I've seen a lot of procedures. Most of them produce muck, and they don't seem to yield very well.

Oliver said it pretty well: https://www.youtube.com/watch?v=upD6cB9Rzvk

[Edited on 16-5-2018 by zed]
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[*] posted on 16-5-2018 at 00:41


Can anyone write out the dimensional analysis for me.

30grams KI, how much H2O2 needed/HCL.
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[*] posted on 16-5-2018 at 00:52


Does it have to be that specific problem, or can it just be something similar?

You usually convert solutions to molarity from percents before you use them, so as to better estimate moles. In that system, I believe you can then calculate moles by multiplying the volume in liters by the molarity.

There are also "normality" solutions, which is different than "molarity", but I'm not sure I remember how that works well enough to explain it.

Using a solution of a known molarity makes it really easy to figure out the moles based on volume. So to answer your question, you'd probably compute the molarity of the solution and then write it on the bottle.




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[*] posted on 16-5-2018 at 06:58


I'll second the recommendation to not use this particular reaction if you're after iodine. Both iodide and chloride ions are catalysts for decomposing hydrogen peroxide, so most of your oxidizer will be wasted to that and your yields will be very low. I actually had a hard time finding an equation for this method because of that, but I think this is it:
2HCl + 2KI + H2O2 ⟶ I2 + 2KCl + 2H2O

While this isn't a great reaction to use, let's answer your math question anyway to help build these skills back up. We'll calculate things using 30g KI.

Molecular Weights (in g/mol)
HCl - 36.5
KI - 166
H2O2 - 34
I2 - 253.8

Hydrogen Peroxide
First, it's helpful to convert this to molarity to make calculation easier.
Assuming this is a weight/volume measure, 3% means 3g peroxide in 100mL solution. But we need to know the volume of water, not the resulting solution, so we divide by the density: (100mL) / (1.0095 g/mL) = 99mL
(Since the density of 3% peroxide is so close to 1, we could just use 100mL in our calculations without worrying too much.)
To find molarity,
H2O2 molarity.jpg - 17kB
Thus, our 3% peroxide is 0.89 M.

With that, we can calculate the rest.
H2O2.jpg - 35kB

Hydrochloric Acid
You need to know the concentration of your acid to do this calculation, so I assumed 9.5M. That's the concentration of my hardware store muriatic acid. You can substitute whatever number you have for yours.
HCl.jpg - 33kB

Iodine
I2.jpg - 27kB

Summary
So, assuming I did my math right, for 30g KI you will need 101.5 mL 3% H2O2 and 19 mL 9.5M HCl to produce 22.9 g I2.

But again, your actual yield will be far less than this due to decomposition of the peroxide from I- and Cl-. You'll need to add quite a lot more peroxide to get the most iodine possible.
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[*] posted on 16-5-2018 at 07:05


I don't know what you guys are talking about, it's a great reaction. Peroxide is so cheap that even if you had to use twice as much as stoichiometry would require, it wouldn't be a problem. I used this method on a multiple hundreds of grams scale a few years ago when I rescued a lot of old, partially oxidized sodium iodide that was going to be thrown out by my high school. I'd be a lot more concerned about the chlorine method. Sounds like a recipe for interhalogens.



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[*] posted on 16-5-2018 at 11:35


Quote: Originally posted by Melgar  
Does it have to be that specific problem, or can it just be something similar?

You usually convert solutions to molarity from percents before you use them, so as to better estimate moles. In that system, I believe you can then calculate moles by multiplying the volume in liters by the molarity.

There are also "normality" solutions, which is different than "molarity", but I'm not sure I remember how that works well enough to explain it.

Using a solution of a known molarity makes it really easy to figure out the moles based on volume. So to answer your question, you'd probably compute the molarity of the solution and then write it on the bottle.


It could be something similar, hmm...
Not sure what else forms precipitate with the 3% hydrogen peroxide though, for instance.

I may be adding further complexity to my question by responding to yours.
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[*] posted on 16-5-2018 at 13:57


The bubbles would always make it so my precipitate never precipitated. With HCl and bleach, on the other hand, I think the interhalogens don't precipitate and are green-colored. So immediately when you notice that color, the reaction is done. After filtering and drying it, it can be purified by sublimation, and it's seemed fine when I've done that.



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[*] posted on 22-8-2020 at 15:52


The chemistry question still remains, because I want to learn chemistry-


in addition to a few other questions,


1) Is it better to do the reaction in hot or cold conditions? Is there an optimal temperature that has ever been worked out, anywhere? (anyone??)


2) Can someone write out the stoichiometry and balanced equation for the reaction between KI, Iodine and Hydrogen Peroxide?


Show how much I2! is produced from an arbitrary amount of KI.


3) How much iodine is in 100grams of KI? --


For this I need to find the atomic mass of Potassium, and that of Iodine then find out what percentages each comprises of the total?


Potassium (K), has an atomic mass of 39.0983 whereas Iodine (I), has an atomic mass of 126.90447, the two together equal 166.0027,


4) How many significant figures get rounded here?


5) The iodine comprises 7644723248% of the compound, and the


potassium comprises the rest, -- again, can someone help me with the


significant figures here in the calculation? Significant figures have still


eluded me, additionally, by the number 7644723248%, I mean to write


.7644723248, how is this expressed?


I think it is a good reaction, it is viable for producing iodine, and it brings up


a good chemistry question.


Thanx









[Edited on 8/22/2020 by Yttrium2]
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[*] posted on 22-8-2020 at 17:59


Yttrium2 this is how I will approach this if I wanted to do it being a rank amateur. I’ll put the equation into my app “Stoichiometry” and see if the app can balance it; in this case it comes out as mention higher up in this thread:
2KI + 2HCl + H2O2 = I2 + 2KCl +2HO
Then i enter my known reagent (or product) mass, in this case I entered 20g for KI. The app then spits out the other masses, in this case, all in gram:

KI - 20
HCl - 4.4
H2O2- 2
I2 - 15.3
KCl - 9
H2O - 2.2

KI is easy just measure out 20g

HCl at 32% contraction density is close enough to 1 just assume it is 1 and thus to calculate the weight of 32% HCl to use it is (4.4/32)*100 = 13.75g

Similarly 3% H2O2 will be (2/3)*100 = 66g

Add extra HCl and H2O2 as required based on what you see or read up beforehand.

For my, with my balance, techniques, and aims no point to think about more than 1 decimal when working with grams.

Any obvious errors with this practical approach please tell me so I can also learn!


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[*] posted on 9-10-2020 at 08:01


My first post! I tried this KI + HCl + H2O2 reaction to make some iodine crystals for my element collection. It seems like it takes quite a while, so I let it sit overnight in the garage. The next day I filtered out a small amount of iodine, which dried into nice sparkly crystal powder. To the liquid, I added some more HCl and 3% H2O2, and let it sit another night. This time, there is a bunch of white precipitate settled on top of a small amount of iodine at the bottom. The solution is pretty dilute at this point, so I was surprised to see anything but iodine precipitate out. Have I made KIO4?
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