11th Class Chemistry Some Basic Concepts of Chemistry

  • 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.


    \[\underset{\underset{1\times \,28g}{\mathop{1mol}}\,}{\mathop{{{N}_{2}}(g)}}\,+\underset{\underset{3\,\times \,\,2g}{\mathop{3mol}}\,}{\mathop{3{{H}_{2}}(g)}}\,\to \underset{\underset{2\times 17g}{\mathop{2mol}}\,}{\mathop{2N{{H}_{3}}(g)}}\,\] 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.

You need to login to perform this action.
You will be redirected in 3 sec spinner