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Reaction between K2O and HCl


Primarygun

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the K2O would be the more reactive.

 

it will explode on contact with water, you can well imagine it`s reaction with an Acid!

you`ll not notice too much of difference between the Na and K, but K is the more reactive of the 2.

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K2O(s) is really so dangerous with HCl?

My book said the rate of KOH(aq)+HCl and NaOH(aq) + HCl is more or less the same, is it correct?

I have some anhydrous powder. It is a calcium compound.

How I can distinguish calcium oxide and calcium hydroxide?

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Then' date=' it is more reactive than Cu(OH)2?

But to get K by reducing, electrolysis is needed, why? but to get Cu, carbon is needed only.[/quote']

 

The bonds may be weak, but K is very reactive and hence will react with OH in water as soon as it has been electrolysed.

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Actually, if memory serves me it is next to impossible to make or obtain anhydrous KOH. I don't actually remember why, but I will look it up. (I seem to remember something about it getting trapped in the crystal structure.)

 

I also remember something about the K - OH bond being more stable in a weaker bonding state than what you would calculate for Na. If I'm remembering this right, it has to do with its so-called quantum bonding orbital.

That is, the more stable configuration is somewhat more energetic, thus somewhat easier to break.

 

Darn. I need to look this up.

 

Addendum - Well, I do see listings for "Anhydrous KOH", but I was almost certain that I learned somewhere along the way that it is virtually impossible to get all the water out of KOH. Maybe I'm wrong about this. (Wouldn't be the first time = :D ).

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KOH is pretty hygroscopic, and I certainly wouldn't suggest electrolyzing the molten salt anyplace but inside an argon drybox. (Since molten potassium will tend to ignite in normal, water/oxygen saturated air). But there is no water inherently bound to KOH. If there was, we'd have to write it as [math]KOH.xH_2O[/math]. Like we do with many hydrated copper salts.

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KOH is deliquescent, mind you, and not fun to get on you. if you have KOH, chances are it is at least slightly hydrated, unless you keep it closely mixed with, say, magnesium perchlorate.

 

in other news, i have a nice burn on my arm from a drop of saturated KOH.

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Budulle,

 

Maybe that is what I am remembering, that it is "practically impossible" to keep dry. But I think I remember more than that, some reason it was so hard to "remove" or perhaps to maintain. (see below).

 

No Dejurg, I dont think that I'm suggesting that the H2O is lattice bonded as with other hydrated molecules as represented with the .H2O suffix. (Actually I've never seen KOH written that way).

 

Actually, ACS grade is only required to be 86% See here KOH ACS Grade .

 

 

Except for when I was a student, I have never worked on any industrial process or curious experimentation that required dry KOH. The usual 15% H2O USP/FCC grade was sufficient.

 

As I said, it is possible that I'm not remembering something correctly.

What I remember is some time ago while co-oping (I was a student then) for a process consulting firm who had a client they had done some work for who had a process that used KOH at some intermediate stage, and the efficiency of the conversation had to do with the purity of industrial KOH, and how it always has some water in it, and that it was practically impossible to get it fully dry. Apparently the driness of the KOH affected the yield of the process.

 

I've been doing some research on it, and haven't found anything definitive yet, save what Budulle is describing.

 

I'll let you know if I find anything, Just consider it my hunch(not necessarily true) until I can get some verification.

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Buddle,

 

OK, it's coming back to me now (geez, you must think I'm some senile old lady). As I remember what they did was to add both a dessicant and some CaO to the part of the process that needed the KOH as the reactant. The dessicant would absorb while the CaO would combine with the water shed by the KOH flakes to form the hydroxide Ca(OH)2.

This helped greatly because the main reaction going on experienced a stearic hindurance from the presence of the water. Thus this in-situ sub-process that effectively removes the H2O(or any other reasonable sub-process) improves the efficiency of the main reaction.

 

Such a sub-process was necessitated because of the virtual impossibility (or perhaps unfeasibility) of making or acquiring very dry KOH.

 

I will still search some general references as I get some time and post if I find anything one way or the other.

 

I probably shouldn't (and have not) give much more specific info on that particular reaction. Although it was a long time ago I'm sure they had me sign an NDA.(Some companies are anal about that stuff)

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Hey Fortuna. Now that you said this was for an industrial process, it all becomes a lot clearer to me. The reason why the KOH they used had water in it was because it's just cheaper that way. You can buy anhydrous KOH, but it has to be packaged so that no water can be absorbed from the atmosphere. As a result, the cost goes WAY up. It's cheaper to just dry it yourself, which is what they used with the CaO/Dessicant. So in your situation, it was just more cost effective to use the readily available, more inexpensive KOH which has already absorbed water. :D

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