A) 1.0 mole of \[{{H}_{2}}O\]is produced
B) 1.0 mole of \[NO\]will be produced
C) All the oxygen will be consumed
D) All the ammonia will be consumed
Correct Answer: C
Solution :
\[\begin{matrix} 4N{{H}_{3(g)}}+5{{O}_{2(g)}}\to 4N{{O}_{(g)}}+6{{H}_{2}}{{O}_{(g)}}\, \\ t=0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \,t=t\,\,\,\,\,\,1-4x\,\,\,\,1-5x\,\,\,\,\,\,\,\,4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{matrix}\] Oxygen is limiting reagent So, \[X=\frac{1}{5}=0.2\] all oxygen consumed Left \[N{{H}_{3}}=1-4\times 0.2=0.2\].
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