question_answer 32)

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation : \[{{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g)\] (i) Calculate the mass of ammonia produced if \[2\times {{10}^{3}}g\] dinitrogen reacts with \[1\times {{10}^{3}}g\] of dihydrogen. (ii) Will any of the two readouts remain unreacted? (iii) If yes, which one and what would be its mass?

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