8 L of H_2 and 6 L of Cl_2 react to the max | vedclass

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(B) The balanced chemical equation is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$.According to Avogadro's Law,at constant $P$ and $T$,the volume ratio is

Vedclass · Accepted Answer

$14$

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(B) The balanced chemical equation is: $H_2(g) + Cl_2(g) ightarrow 2HCl(g)$.According to Avogadro's Law,at constant $P$ and $T$,the volume ratio is the same as the mole ratio.Initial volumes: $V(H_2) = 8 \ L$,$V(Cl_2) = 6 \ L$,$V(HCl) = 0 \ L$.Since $Cl_2$ is the limiting reagent,it will be completely consumed.Volume of $Cl_2$ consumed = $6 \ L$.Volume of $H_2$ consumed = $6 \ L$ (as per $1:1$ stoichiometry).Volume of $HCl$ produced = $2 	imes 6 \ L = 12 \ L$.Remaining volume of $H_2 = 8 \ L - 6 \ L = 2 \ L$.Final volume of the mixture = $V(H_2)_{remaining} + V(HCl)_{produced} = 2 \ L + 12 \ L = 14 \ L$.

$8 \ L$ of $H_2$ and $6 \ L$ of $Cl_2$ react to the maximum extent. Find the final volume of the reaction mixture in $L$. Assume $P$ and $T$ remain constant throughout the process.

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