Mega's procedure for making HMX, do I really have to use 95% + NA?

[Dhzugasvili]

So, what happens if I don't use such highly concentrated NA? Any of you who are knowledgeable in this procedure please tell me if it is possible to use say...85% NA?

[PHILOU Zrealone]

Explosion due to a runaway or only hexamine dinitrate salt... no RDX nor HMX.

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"Life that deadly disease sexually transmitted".
"Chemistry is all what stinks and explode; Physic is all what never works! ;-p :-) :o)"

[Detonator]

Philo do you have any info regarding hexamine dinitrate? VoD..etc?

[PHILOU Zrealone]

VOD is arround 7000m/s! And there is a tread about this and about R-salt where all this was extensively discussed within the past two month.

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"Life that deadly disease sexually transmitted".
"Chemistry is all what stinks and explode; Physic is all what never works! ;-p :-) :o)"

[Detonator]

Shady said:
Foss,Hirst,Jones,Springall,A.T Thomas and T.Urbanski have established that boiling an aqueous solution of hexamine dinitate causes the substance to transform into methylhexamethylene mononitrate

does it mean that you cant boil the water out of it?

[Dhzugasvili]

Since I already have highly concentrated NA all the above and the title of this topic is bullshit. NOW i have another question regarding Mega's procedure for making HMX. IT SAYS THAT THE YIELD IS 95%, what exactly does this mean? I think I have asked this question before, but I shall ask again, Such a fuckload of reactant volumes, how many grams of HMX does one end up with?? I ask this so I can calculate the volumes with regard to the end product mass. I don't want to end up with a ridiculous amount of the stuff! If anyone has knowledge to share with me on mega's procedure for hmx and how much HMX one ends up with following the recipe exactly, please, it will be much appreciated.

[PHILOU Zrealone]

95% yield means that you will get 95% of the theorical stoechiometric yield. This is based on the hexamine because you use an excess HNO3 of course to fully nitrate the amino parts.
Ex (pure fiction just as example):
(CH2NH)4 + 4HNO3 (in excess in reality)--> (CH2NNO2)4 + 4H2O
THus a 95% yield would mean that starting from one mole of (CH2NH)4 , you get 0.95 mole of (CH2NNO2)4 . Of course when you know this you have to calculate the molar mass of the concerned molecules:
MM(CH2NH)4 is 29*4= 116g/mole
MMHNO3 is 63g/mole
MM(CH2NNO2)4 is 74*4=296g/mole
This means following the equation that 1 mole of (CH2NH)4 react with four of HNO3 to yield (if the reaction was 100% perfect) 1 mole of (CH2NNO2)4; or that 116g tetramethyleneimine reacts with 252g (100% pure HNO3- don't care about this since you use a large excess of it) to produce a theorical yield of 252g of HMX. If the real yield is 0.95 then 252g*0.95= 239.4g.
You notice that 95% is lower than 100% yield, but that the yield in gram is double as compared to the starting 116g of amine.

I hope you have understood, if not I can't help you more.


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"Life that deadly disease sexually transmitted".
"Chemistry is all what stinks and explode; Physic is all what never works! ;-p :-) :o)"