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

Jul 25, 2017

We need $\text{3 equiv}$ of $\text{nitric acid}$ per equiv of $\text{ammonia/ammonium}$.

#### Explanation:

$\text{Reduction half equation:}$

$H N {O}_{3} \left(a q\right) + {H}^{+} + {e}^{-} \rightarrow N {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$
$\stackrel{+ V}{N} \rightarrow \stackrel{+ I V}{N}$

$\text{Oxidation half equation:}$

$N {H}_{4}^{+} \rightarrow \frac{1}{2} {N}_{2} \left(g\right) \uparrow + 3 {e}^{-} + 4 {H}^{+}$
$\stackrel{- I I I}{N} \rightarrow \stackrel{0}{N}$

$3 H N {O}_{3} \left(a q\right) + N {H}_{4}^{+} \rightarrow 3 N {O}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right) + \frac{1}{2} {N}_{2} \left(g\right) \uparrow + {H}^{+}$

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.....

$3 H N {O}_{3} \left(a q\right) + N {H}_{3} \left(a q\right) \rightarrow 3 N {O}_{2} \left(g\right) + \frac{1}{2} {N}_{2} \left(g\right) \uparrow + 3 {H}_{2} O \left(l\right)$

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