A) NO
B) \[N{{O}_{2}}\]
C) \[N{{H}_{4}}N{{O}_{3}}\]
D) \[ZnN{{O}_{3}}\]
Correct Answer: C
Solution :
\[Zn+10HNO{{ & }_{3}}\xrightarrow{{}}4Zn{{(N{{O}_{3}})}_{2}}+\underset{\begin{smallmatrix} \text{Ammonium}\, \\ \,\,\,\,\,\,\,\,\,\text{nitrate} \end{smallmatrix}}{\mathop{N{{H}_{4}}NO{{ & }_{3}}}}\,\] \[+\,3{{H}_{2}}O\] \[\therefore \] Zn reacts with cold dil. \[HN{{O}_{3}}\]to produce \[N{{H}_{4}}N{{O}_{3}}.\] With dil. \[\text{HN}{{\text{O}}_{\text{3}}}\]it produces\[-{{\text{N}}_{\text{2}}}\text{O}\] (nitrous oxide) With conc. \[HN{{O}_{3}}\]it produces \[-N{{O}_{2}}\](nitrogen dioxide)
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