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tom haggen
June 23rd, 2005, 10:07 PM
Is it even possible to balance an equation for the reaction between hexamethylenetetramine and nitric acid to form RDX?


To start with you have;

C6H12N4 + HNO3 <====> C3H6N6O6 + ?

This equation was hard for me to balance for some reason. So I calculated moles from megalomania's method and found that he was using a 16:1 ratio in mols of HNO3 and C6H12N4. So I tried balancing that and it didn't work. I assumed that he was just recommending excess nitric acid to compensate for everyone. I also assume that the by product of this reaction is good ol’ H2O. I just can't balance it for the life of me.

tom haggen
June 23rd, 2005, 10:07 PM
Is it even possible to balance an equation for the reaction between hexamethylenetetramine and nitric acid to form RDX?


To start with you have;

C6H12N4 + HNO3 <====> C3H6N6O6 + ?

This equation was hard for me to balance for some reason. So I calculated moles from megalomania's method and found that he was using a 16:1 ratio in mols of HNO3 and C6H12N4. So I tried balancing that and it didn't work. I assumed that he was just recommending excess nitric acid to compensate for everyone. I also assume that the by product of this reaction is good ol’ H2O. I just can't balance it for the life of me.

cyclonite4
June 23rd, 2005, 10:19 PM
Yes, it is quite possible to balance the equation.
This seems right to me (and is consistant with reports of ammonia and formaldehyde being released):

C6H12N4 + 3HNO3 <===> (CH2)3(NNO2)3 + 3CH2O + NH3

With various methods of synthesis, HNO3 is kept heavily in excess (equilibrium thing). There is also no production of water by the reaction (water hinders the reaction).

cyclonite4
June 23rd, 2005, 10:19 PM
Yes, it is quite possible to balance the equation.
This seems right to me (and is consistant with reports of ammonia and formaldehyde being released):

C6H12N4 + 3HNO3 <===> (CH2)3(NNO2)3 + 3CH2O + NH3

With various methods of synthesis, HNO3 is kept heavily in excess (equilibrium thing). There is also no production of water by the reaction (water hinders the reaction).

tom haggen
June 24th, 2005, 01:22 AM
I guess that’s the part I was having some trouble with was being absolutely sure of the products of the reaction. Also since this is a substitution reaction, I didn't really know if you were supposed to balance it in a linear fashion. Sticking to megalomania's method would probably be the best idea; it would just be nice to be able to predict yields with variant amounts of reactants.

tom haggen
June 24th, 2005, 01:22 AM
I guess that’s the part I was having some trouble with was being absolutely sure of the products of the reaction. Also since this is a substitution reaction, I didn't really know if you were supposed to balance it in a linear fashion. Sticking to megalomania's method would probably be the best idea; it would just be nice to be able to predict yields with variant amounts of reactants.