Referring to the following reactions the missing products A,B,C and D respectively are :
$\begin{array}{l}{\mathrm{NH}}_4\mathrm{Cl}\text{ (aq.) }+{\mathrm{NaNO}}_2\text{ (aq.) }⟶\left[\mathrm{A}\right]+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+\mathrm{NaCl}\text{ (aq.) }\\ {\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\stackrel{\mathrm{\Delta }}{⟶}\left[\mathrm{B}\right]+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{Cr}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)\\ \mathrm{Cu}+{\mathrm{HNO}}_3\text{ (dilute) }⟶\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+\left[\mathrm{C}\right]+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ \mathrm{Cu}+{\mathrm{HNO}}_3\text{ (concentrate) }⟶\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+\left[\mathrm{D}\right]+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\end{array}$
 

In the laboratory dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
${\mathrm{NH}}_4\mathrm{Cl}(\mathrm{aq}.)+{\mathrm{NaNO}}_2(\mathrm{aq}.)⟶{\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+\mathrm{NaCl}(\mathrm{aq}.)$
Small amounts of NO and HNO₃ are also formed as impurities, which can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.
(ii) Dinirogen can also be prepared by heating ammonium dichromate.
${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{⟶}{\mathrm{N}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{Cr}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$
(iii) Copper metal reacts with dilute HNO₃ to form nitrogen (II) oxide (NO).
$3\mathrm{Cu}\left(\mathrm{s}\right)+8{\mathrm{HNO}}_3\text{ (dilute) }⟶3\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2(\mathrm{aq}.)+2\mathrm{NO}\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
(iv) Copper metal reacts with conc. HNO₃ to form nitrogen (IV) oxide or nitrogen dioxide (NO₂).