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