A) \[\frac{1}{6}\]
B) \[\frac{5}{6}\]
C) \[\frac{1}{3}\]
D) none of these
Correct Answer: D
Solution :
| \[\underset{1}{\mathop{{{N}_{2}}}}\,+\underset{4}{\mathop{3{{H}_{2}}}}\,\xrightarrow{{}}\underset{0}{\mathop{2N{{H}_{3}}}}\,\] |
| \[N{{H}_{3}}+{{H}_{2}}O\xrightarrow{{}}N{{H}_{4}}OH\] |
| \[\underset{1\,\,\,mole}{\mathop{N{{H}_{4}}OH+HCl}}\,\xrightarrow{{}}\underset{1\,\,mole}{\mathop{N{{H}_{4}}Cl+{{H}_{2}}O}}\,\] |
| So \[N{{H}_{3}}\]formed is 1 mole |
| \[{{N}_{2}}+3{{H}_{2}}\xrightarrow{{}}2N{{H}_{3}}\] |
| \[\left( 1-\frac{1}{2} \right)\,\,\,\left( 4-\frac{3}{2} \right)\,\,\,\,\,\,\,\,\,(1)\] |
| \[\begin{matrix} \frac{1}{2}, & \frac{5}{2} & 1 \\ \end{matrix}\] |
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