For the reaction,N_{2(g)} + 3H_{2(g)} to 2NH_ | vedclass

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(B) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$.According to the stoichiometry,$1 \ mole$ of $N_2$ $(28 \ g)$ reacts with

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$56 \ g$ of $N_2 + 10 \ g$ of $H_2$

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(B) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$.According to the stoichiometry,$1 \ mole$ of $N_2$ $(28 \ g)$ reacts with $3 \ moles$ of $H_2$ $(6 \ g)$.For option $A$: $14 \ g$ of $N_2$ $(0.5 \ mole)$ requires $1.5 \ moles$ of $H_2$ $(3 \ g)$. Since $4 \ g$ of $H_2$ is present,$H_2$ is in excess.For option $B$: $56 \ g$ of $N_2$ $(2 \ moles)$ requires $6 \ moles$ of $H_2$ $(12 \ g)$. Since only $10 \ g$ of $H_2$ $(5 \ moles)$ is available,$H_2$ is the limiting reagent.For option $C$: $28 \ g$ of $N_2$ $(1 \ mole)$ requires $3 \ moles$ of $H_2$ $(6 \ g)$. Since $6 \ g$ of $H_2$ is present,it is a stoichiometric mixture.For option $D$: $35 \ g$ of $N_2$ $(1.25 \ moles)$ requires $3.75 \ moles$ of $H_2$ $(7.5 \ g)$. Since $8 \ g$ of $H_2$ is present,$H_2$ is in excess.

For the reaction,$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$,identify dihydrogen $(H_2)$ as the limiting reagent in the following reaction mixtures.

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