$\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{H}_{2} \mathrm{NO}_{3}^{+}+\mathrm{HSO}_{4}^{-}$
$\mathrm{H}_{2} \mathrm{NO}_{3}^{+} \rightarrow \mathrm{NO}_{2}^{+}+\mathrm{H}_{2} \mathrm{O}$
Hence, in this reaction HNO $_{3}$ acts as a base and $\mathrm{H}_{2} \mathrm{SO}_{4}$ as an acid.
(image) $\xrightarrow{{[O]}}A\xrightarrow{{SOC{l_2}}}B\xrightarrow{{Na{N_3}}}C\xrightarrow{{Heat}}D$