In the reaction N_2 + 3H_2 rightarrow 2NH_3,t | vedclass

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(D) According to $Gay-Lussac's$ Law of Gaseous Volumes,when gases react together,they do so in volumes which bear a simple whole number ratio to one a

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Gay-Lussac's Law of Gaseous Volumes

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(D) According to $Gay-Lussac's$ Law of Gaseous Volumes,when gases react together,they do so in volumes which bear a simple whole number ratio to one another and to the volume of the products,if gaseous,provided that the temperature and pressure remain constant. In the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$,the stoichiometric coefficients are $1$,$3$,and $2$ respectively. Since the reactants and products are in the gaseous state,the ratio of their volumes is $1 : 3 : 2$,which is a simple whole number ratio. Therefore,this illustrates $Gay-Lussac's$ Law of Gaseous Volumes.

In the reaction $N_2 + 3H_2 \rightarrow 2NH_3$,the ratio of volumes of $N_2$,$H_2$,and $NH_3$ is $1 : 3 : 2$. This illustrates which law?

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