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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