18 L mixture of N_2 and H_2 gives maximum 6 | vedclass

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(D) The balanced chemical equation for the formation of ammonia is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ According to Avogadro's Law,at constant te

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$1 : 3$

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(D) The balanced chemical equation for the formation of ammonia is: $N_2(g) + 3H_2(g) ightarrow 2NH_3(g)$ According to Avogadro's Law,at constant temperature and pressure,the volume of gases is proportional to the number of moles. From the stoichiometry,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$. To produce $6 \ L$ of $NH_3$,we require: $V(N_2) = \frac{1}{2} 	imes 6 \ L = 3 \ L$ $V(H_2) = \frac{3}{2} 	imes 6 \ L = 9 \ L$ Total volume of $N_2$ and $H_2$ used = $3 \ L + 9 \ L = 12 \ L$. However,the initial mixture is $18 \ L$. This implies there is an excess of $6 \ L$ of one of the gases. If $N_2$ is in excess,initial $N_2 = 3 + 6 = 9 \ L$ and $H_2 = 9 \ L$. Ratio $N_2 : H_2 = 9 : 9 = 1 : 1$. If $H_2$ is in excess,initial $H_2 = 9 + 6 = 15 \ L$ and $N_2 = 3 \ L$. Ratio $N_2 : H_2 = 3 : 15 = 1 : 5$. Both ratios are possible,so the answer is $(A)$ and $(B)$ both.

$18 \ L$ mixture of $N_2$ and $H_2$ gives maximum $6 \ L$ of $NH_3$ at same temperature and pressure. What will be the ratio of $N_2$ and $H_2$ initially taken?

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