Question

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

Answer 1

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.