Answer:
(i) The given reaction is:
\[\underset{\underset{28g}{\mathop{1mol}}\,}{\mathop{{{N}_{2}}(g)}}\,+\underset{\underset{6g}{\mathop{3\,mol}}\,}{\mathop{3{{H}_{2}}(g)}}\,\to
\underset{\underset{34g}{\mathop{2\,mol}}\,}{\mathop{2N{{H}_{3}}(g)}}\,\]
Case I: Let
\[{{\mathbf{N}}_{\mathbf{2}}}\] is completely consumed:
\[\because
\]\[28g{{N}_{2}}\]gives \[34g\,N{{H}_{3}}\]
\[\therefore \]
\[2000g\,{{N}_{2}}\]will give
\[\frac{34}{28}\times
2000g\,N{{H}_{3}},i.e.5666.6g\,N{{H}_{3}}\], i.e.
\[2428.5gN{{H}_{3}}\]
Case II: Let
\[{{\mathbf{H}}_{\mathbf{2}}}\] is completely consumed:
\[\because \]6g\[{{H}_{2}}\]
gives 34 g \[N{{H}_{3}}\]
\[\therefore
\]1000 g \[{{H}_{2}}\] will give \[\frac{34}{6}\times 1000gN{{H}_{3}}\], i.e.,
\[5666.6g\,N{{H}_{3}}\]
Since, \[{{N}_{2}}\]
gives least amount of product on complete consumption hence it will be limiting
reagent and amount of ammonia formed will be 2428.5 g.
(ii) \[{{H}_{2}}\]will
be excess reactant.
(iii) \[\underset{\begin{smallmatrix}
\text{Before}\,\text{reaction}
\\
\text{After}\,\text{reaction}
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\underset{0}{\mathop{2000g}}\,}{\mathop{{{N}_{2}}(g)}}\,+\underset{\underset{x}{\mathop{1000g}}\,}{\mathop{3{{H}_{2}}(g)}}\,\to
\underset{\underset{2428.5g}{\mathop{0}}\,}{\mathop{2N{{H}_{3}}(g)}}\,\]
According to
the law of conservation of mass,
\[2000\text{ }+\text{ }1000\text{ }=x+\text{ }2428.5\]
\[\therefore
\] \[x=571.5g\]
Mass of \[{{H}_{2}}\]
remaining = 571.5 g
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