• # 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

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