When 22.4 L of H_{2(g)} is mixed with 11.2 | vedclass

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(A) The reaction is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$At $S.T.P.$,$22.4 \ L$ of any gas corresponds to $1 \ mol$.Initial moles of $H_2 = \

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$1 \ mol$ of $HCl_{(g)}$

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(A) The reaction is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$At $S.T.P.$,$22.4 \ L$ of any gas corresponds to $1 \ mol$.Initial moles of $H_2 = \frac{22.4 \ L}{22.4 \ L/mol} = 1 \ mol$.Initial moles of $Cl_2 = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.According to the stoichiometry,$1 \ mol$ of $H_2$ reacts with $1 \ mol$ of $Cl_2$.Since $Cl_2$ is present in a smaller amount $(0.5 \ mol The amount of $HCl$ formed is determined by the limiting reagent $(Cl_2)$:$1 \ mol$ of $Cl_2$ produces $2 \ mol$ of $HCl$.Therefore,$0.5 \ mol$ of $Cl_2$ produces $2 	imes 0.5 = 1.0 \ mol$ of $HCl$.

When $22.4 \ L$ of $H_{2(g)}$ is mixed with $11.2 \ L$ of $Cl_{2(g)}$,each at $S.T.P.$,the moles of $HCl_{(g)}$ formed is equal to

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