To solve this problem, we need to determine the moles of $$HCl (g)$$ formed when $$H$_2$ and $$Cl$_2$ are mixed.$\\ \\$The balanced chemical equation for the reaction is:$\\$H$_2(g) +$Cl$_2(g) \rightarrow 2$HCl$(g) \\ \\$At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.$\\ \\$Given:$\\$• Volume of $$H$_2 = 22.4$ L$\\$• Volume of $$Cl$_2 = 11.2$ L$\\ \\$Calculate the moles of each gas:$\\$Moles of $H_2 = \frac{22.4 \ L}{22.4 \ L/mol} = 1$ mol$\\$Moles of ${Cl}_2 = \frac{11.2 L}{22.4 \ L/mol} = 0.5$ mol$\\ \\$According to the balanced equation, 1 mole of $$H$_2$ reacts with 1 mole of $$Cl$_2$ to produce 2 moles of $$HCl$. \\ \\$Since $$Cl$_2$ is the limiting reagent (0.5 mol), it will determine the amount of $$HCl$$ formed.$\\ \\$Using the stoichiometry of the reaction:$\\ 0.5$ mol of $$Cl$_2 \rightarrow 2 \times 0.5$ mol of $$HCl$= 1$ mol of $$HCl$\\ \\$Therefore, the moles of $$HCl$$ formed is $1$ mol.$\\ \\$This corresponds to Option 1.