How many equiv of nitric acid are required to oxidize one equiv of ammonia to give an half an equiv of dinitrogen gas?

1 Answer
anor277
Jul 25, 2017

We need "3 equiv" of "nitric acid" per equiv of "ammonia/ammonium".

Explanation:

"Reduction half equation:"

HNO_3(aq) +H^(+) + e^(-) rarr NO_2(g) +H_2O(l)
stackrel(+V)Nrarrstackrel(+IV)N

"Oxidation half equation:"

NH_4^(+) rarr 1/2N_2(g)uarr +3e^(-) + 4H^+
stackrel(-III)Nrarrstackrel(0)N

3HNO_3(aq) +NH_4^(+)rarr 3NO_2(g) +3H_2O(l)+1/2N_2(g)uarr + H^+

And we can make it a bit simpler than this....by representing the oxidation of ammonia rather than ammonium.....and so we remove H^+ from each side of the equation.....

3HNO_3(aq) +NH_3(aq)rarr 3NO_2(g) +1/2N_2(g)uarr+3H_2O(l)

The which is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical reality.

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