${\mathrm{CN}}^{-}$ is oxidised by ${\mathrm{NO}}_3^{-}$ in presence of acid :

$a{\mathrm{CN}}^{-}+b{\mathrm{NO}}_3^{-}+c{\mathrm{H}}^{+}⟶(a+b)\mathrm{NO}+a{\mathrm{CO}}_2+\frac{c}{2}{\mathrm{H}}_2\mathrm{O}$

What are the values of a, b, c in that order :

$a{\mathrm{CN}}^{-}+b{\mathrm{NO}}_3^{-}+{\mathrm{CH}}^{+}\to (a+b)\mathrm{NO}+a{\mathrm{CO}}_2+\frac{c}{2{\mathrm{H}}_2\mathrm{O}}$

Ion Electron method (Oxidation Half )

$\begin{array}{l}3\left\{3{\mathrm{H}}_2\mathrm{O}+{\mathrm{CN}}^{+2-3}\to {\mathrm{NO}}^{+2-2}+{\mathrm{CO}}_2^{-2}+6{\mathrm{H}}^{+}+7e^{-}\right.\\ \left.3e^{-}+4{\mathrm{H}}^{+}+{\mathrm{NO}}_3^{-2}\to +{\mathrm{NO}}^{-2}+2{\mathrm{H}}_2\mathrm{O}\text{ (reduction half) }\right\}7\\ \Rightarrow 3{\mathrm{CN}}^{-}+7{\mathrm{NO}}_3^{-}=10\mathrm{NO}+3{\mathrm{CO}}_2+5{\mathrm{H}}_2\mathrm{O}\\ \frac{c}{2}=5\Rightarrow c=10\\ \therefore (a,b,c)\equiv (3,7,10)\end{array}$