When $11.2 \mathrm{~L}$ of $\mathrm{H}_{2(g)}$ is mixed with $33.6 \mathrm{~L}$ of $\mathrm{Cl}_{2(g

Question

When $11.2 \mathrm{~L}$ of $\mathrm{H}_{2(g)}$ is mixed with $33.6 \mathrm{~L}$ of $\mathrm{Cl}_{2(g)}$, each measured at STP, the total number of moles of $\mathrm{HCl}_{(g)}$ formed upon complete reaction is

Professor PC Thomas Classes · Answer

(a) : At STP, 1 mole of any ideal gas occupies $22.4 \mathrm{~L}$. Moles of $\mathrm{H}_2 = \frac{11.2}{22.4} = 0.5 \mathrm{~mol}$. Moles of $\mathrm{Cl}_2 = \frac{33.6}{22.4} = 1.5 \mathrm{~mol}$. The balanced chemical equation is $\mathrm{H}_2 + \mathrm{Cl}_2 ightarrow 2\mathrm{HCl}$. From this, 1 mole of $\mathrm{H}_2$ reacts with 1 mole of $\mathrm{Cl}_2$ to produce 2 moles of $\mathrm{HCl}$. Since only $0.5 \mathrm{~mol}$ of $\mathrm{H}_2$ is available, it is the limiting reagent. The moles of $\mathrm{HCl}$ produced will be $2 	imes 0.5 = 1.0 \mathrm{~mol}$.