A) \[N{{H}_{4}}Cl\]
B) \[{{N}_{2}}+HCl\]
C) \[{{N}_{2}}+N{{H}_{4}}Cl\]
D) \[{{N}_{2}}+NC{{l}_{3}}\]
E) \[NC{{l}_{3}}+HCl\]
Correct Answer: C
Solution :
\[3C{{l}_{2}}+\underset{\text{excess}}{\mathop{8N{{H}_{3}}}}\,\xrightarrow{{}}6N{{H}_{4}}Cl+{{N}_{2}}\uparrow \]
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