In the Haber process,30 L of dihydrogen and | vedclass

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(C) The balanced chemical equation for the Haber process is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.Initially,we have $30 \ L$ of $N_2$ and $3

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$10 \ L$ ammonia,$25 \ L$ nitrogen,$15 \ L$ hydrogen

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(C) The balanced chemical equation for the Haber process is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.Initially,we have $30 \ L$ of $N_2$ and $30 \ L$ of $H_2$.According to the stoichiometry,$1 \ L$ of $N_2$ reacts with $3 \ L$ of $H_2$ to produce $2 \ L$ of $NH_3$.If all $30 \ L$ of $H_2$ were consumed,it would require $10 \ L$ of $N_2$ and produce $20 \ L$ of $NH_3$ (theoretical yield).Since the reaction yielded only $50\%$ of the expected product,the actual amount of $NH_3$ produced is $50\% 	imes 20 \ L = 10 \ L$.For $10 \ L$ of $NH_3$ to be produced,$5 \ L$ of $N_2$ and $15 \ L$ of $H_2$ must have reacted.Remaining $N_2 = 30 \ L - 5 \ L = 25 \ L$.Remaining $H_2 = 30 \ L - 15 \ L = 15 \ L$.Thus,the final composition is $10 \ L$ ammonia,$25 \ L$ nitrogen,and $15 \ L$ hydrogen.

In the Haber process,$30 \ L$ of dihydrogen and $30 \ L$ of dinitrogen were taken for the reaction,which yielded only $50\%$ of the expected product. What will be the composition of the gaseous mixture under the aforesaid condition in the end?

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