When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in  equal volumes according to the equation

$\mathrm{xCu}+\mathrm{y}{\mathrm{HNO}}_3⟶\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+\mathrm{NO}+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

The coefficient x and y are

The balanced equation is 

$\begin{array}{l}\mathrm{Cu}\stackrel{ }{\to }{\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{- } ............. \left(\mathrm{i}\right)\\ {\mathrm{H}}^{\oplus }+{\mathrm{e}}^{-}+{\mathrm{HNO}}_3⟶{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O} ................. \left(\mathrm{ii}\right)\\ 3{\mathrm{H}}^{\oplus }+3{\mathrm{e}}^{-}+{\mathrm{HNO}}_3⟶\mathrm{NO}+2{\mathrm{H}}_2\mathrm{O} ................ \left(\mathrm{iii}\right)\end{array}$

 Multiplying Eq. (i) by 2 and adding Eqs. (i), (ii) and (iii), we get

$$\begin{gathered} 2\mathrm{Cu}+4{\mathrm{H}}^{\oplus }+2{\mathrm{HNO}}_3\stackrel{ }{\to }2{\mathrm{Cu}}^{2+}+{\mathrm{NO}}_2+\mathrm{NO}+3{\mathrm{H}}_2\mathrm{O} \\ \mathrm{or} \\ 2\mathrm{Cu}+6{\mathrm{HNO}}_3⟶2\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{NO}}_2+\mathrm{NO}+3{\mathrm{H}}_2\mathrm{O} \\ \therefore \mathrm{x} \mathrm{and} \mathrm{y} \mathrm{are} 2 \mathrm{and} 6 \end{gathered}$$