When 12 L of H_2 and 11.2 L of Cl_2 are mix | vedclass

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(C) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$.According to the stoichiometry,$1 \ volume$ of $H_2$ reacts with

Vedclass · Accepted Answer

$0.8 \ L \ H_2$ and $22.4 \ L \ HCl_{(g)}$

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(C) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$.According to the stoichiometry,$1 \ volume$ of $H_2$ reacts with $1 \ volume$ of $Cl_2$ to produce $2 \ volumes$ of $HCl_{(g)}$.Given: $V(H_2) = 12 \ L$ and $V(Cl_2) = 11.2 \ L$.Since $Cl_2$ is the limiting reagent $(11.2 \ L $11.2 \ L$ of $Cl_2$ will react with $11.2 \ L$ of $H_2$ to produce $2 	imes 11.2 = 22.4 \ L$ of $HCl_{(g)}$.Remaining $H_2 = 12 \ L - 11.2 \ L = 0.8 \ L$.Thus,the final mixture contains $0.8 \ L$ of $H_2$ and $22.4 \ L$ of $HCl_{(g)}$.

When $12 \ L$ of $H_2$ and $11.2 \ L$ of $Cl_2$ are mixed and allowed to react according to the equation $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$,what is the final composition of the mixture?

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