NH_3 is produced according to the following r | vedclass

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(A) $1$. Identify the limiting reagent: According to the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,$1 \ mol$ of $N_2$ requires $3 \ mol$

Vedclass · Accepted Answer

$75$

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(A) $1$. Identify the limiting reagent: According to the reaction $N_{2(g)} + 3H_{2(g)} ightarrow 2NH_{3(g)}$,$1 \ mol$ of $N_2$ requires $3 \ mol$ of $H_2$.$2$. For $0.5 \ mol$ of $N_2$,the required $H_2$ is $0.5 	imes 3 = 1.5 \ mol$.$3$. Since only $0.5 \ mol$ of $H_2$ is available,$H_2$ is the limiting reagent.$4$. Calculate theoretical yield: $3 \ mol$ of $H_2$ produces $2 \ mol$ of $NH_3$. Therefore,$0.5 \ mol$ of $H_2$ produces $(2/3) 	imes 0.5 = 0.333 \ mol$ of $NH_3$.$5$. Calculate percentage yield: $	ext{Percentage yield} = (	ext{Actual yield} / 	ext{Theoretical yield}) 	imes 100 = (0.25 / 0.333) 	imes 100 \approx 75 \%$.

$NH_3$ is produced according to the following reaction: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$. In an experiment,$0.25 \ mol$ of $NH_3$ is formed when $0.5 \ mol$ of $N_2$ is reacted with $0.5 \ mol$ of $H_2$. What is the percentage yield (in $\%$)?

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