A) one mole of \[N{{H}_{3}}\]
B) two moles of \[N{{H}_{3}}\]
C) one mole of \[HN{{O}_{3}}\]
D) two moles of \[HN{{O}_{3}}\]
Correct Answer: B
Solution :
\[\underset{1\,\,mol}{\mathop{{{\operatorname{Mg}}_{3}}{{N}_{2}}}}\,+\underset{(excess)}{\mathop{6{{H}_{2}}O}}\,\,\,\to \,\,3Mg{{(OH)}_{2}}+\underset{2\,\,mol}{\mathop{2N{{H}_{3}}}}\,\]
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