• 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.