So, assuming this is actually the reaction that is taking place, this would mean I would need 2 moles of NH4NO3 for every 4
moles of Al, since each molecule of AN has 3 Oxygen atoms, so 2 moles of AN would produce 6 moles of atomic oxygen, which would combine into 3 moles
of molecular oxygen. Right?
If so, this means I need 2 moles of Al per 1 mole of AN, or 53.9630772g of Al per 80.0434g of AN, which comes out to be 40.2690% Aluminum, by mass.
This seems like far too much Aluminum; several videos I have seen of Ammonal explosions quote that their mixtures are 90% AN and 10% Al, by mass. Have
I made any errors here, or are they just using non-stoichiometric mixtures?
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