question_answer 5)

\[50kg\,{{N}_{2}}(g)\] and \[10kg\]of \[{{H}_{2}}(g)\] are mixed to produce \[N{{H}_{3}}(g)\]. Calculate the amount of \[N{{H}_{3}}(g)\] formed. Identify the limiting reagent in the production of \[N{{H}_{3}}\] in this situation.

Answer:

\[\underset{\underset{1\times \,28g}{\mathop{1mol}}\,}{\mathop{{{N}_{2}}(g)}}\,+\underset{\underset{3\,\times \,\,2g}{\mathop{3mol}}\,}{\mathop{3{{H}_{2}}(g)}}\,\to \underset{\underset{2\times 17g}{\mathop{2mol}}\,}{\mathop{2N{{H}_{3}}(g)}}\,\] Case I: Let \[{{\mathbf{N}}_{\mathbf{2}}}\] is completely consumed: \[\because \,\,28kg\,{{N}_{2}}\] gives \[34kg\,N{{H}_{3}}\] \[\therefore \]\[50kg\,{{N}_{2}}\] will give \[\frac{34}{28}\times 50kg\,\,N{{H}_{3}}=60.71kg\] Case II: Let \[{{\mathbf{H}}_{\mathbf{2}}}\] is completely consumed: \[\because \]\[6\,kg\,{{H}_{2}}\]gives \[34\,kg\,N{{H}_{3}}\] \[\therefore \]\[10kg\,{{H}_{2}}\] will give \[\frac{34}{6}\times 10kg\,N{{H}_{3}}=56.66Kg\] Since, \[{{H}_{2}}\] gives least amount of product on being completely consumed hence it is limiting reagent and actual amount of ammonia formed will be 56.66 kg.