For the gaseous reaction H_2 + Cl_2 rightarro | vedclass

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(A) According to Gay-Lussac's Law of Gaseous Volumes,the reaction is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$.From the stoichiometry,$1 \ volume$ of $H

Vedclass · Accepted Answer

$40 \ mL, 80 \ mL$

Vedclass · Answer

(A) According to Gay-Lussac's Law of Gaseous Volumes,the reaction is: $H_2(g) + Cl_2(g) ightarrow 2HCl(g)$.From the stoichiometry,$1 \ volume$ of $H_2$ reacts with $1 \ volume$ of $Cl_2$ to produce $2 \ volumes$ of $HCl$.Therefore,$40 \ mL$ of $H_2$ will react with $40 \ mL$ of $Cl_2$ to produce $2 	imes 40 \ mL = 80 \ mL$ of $HCl$.Thus,the volume of chlorine required is $40 \ mL$ and the volume of $HCl$ formed is $80 \ mL$.

For the gaseous reaction $H_2 + Cl_2 \rightarrow 2HCl$,if $40 \ mL$ of hydrogen reacts with chlorine,the volume of chlorine required and the volume of $HCl$ formed are respectively:

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