Aqueous HCl reacts with MnO_{2(s)} to form Mn | vedclass

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(A) The balanced chemical equation for the reaction is: $MnO_{2(s)} + 4HCl_{(aq)} \rightarrow MnCl_{2(aq)} + Cl_{2(g)} + 2H_{2}O_{(l)}$Calculate the m

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$7.1$

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(A) The balanced chemical equation for the reaction is: $MnO_{2(s)} + 4HCl_{(aq)} ightarrow MnCl_{2(aq)} + Cl_{2(g)} + 2H_{2}O_{(l)}$Calculate the molar mass of $MnO_{2}$: $55 + (2 	imes 16) = 87 \ g \ mol^{-1}$.Calculate the number of moles of $MnO_{2}$ used: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{8.7 \ g}{87 \ g \ mol^{-1}} = 0.1 \ mol$.According to the stoichiometry of the reaction, $1 \ mol$ of $MnO_{2}$ produces $1 \ mol$ of $Cl_{2}$.Therefore, $0.1 \ mol$ of $MnO_{2}$ will produce $0.1 \ mol$ of $Cl_{2}$.Calculate the molar mass of $Cl_{2}$: $2 	imes 35.5 = 71 \ g \ mol^{-1}$.Calculate the weight of $Cl_{2}$ liberated: $0.1 \ mol 	imes 71 \ g \ mol^{-1} = 7.1 \ g$.

Aqueous $HCl$ reacts with $MnO_{2(s)}$ to form $MnCl_{2(aq)}$, $Cl_{2(g)}$ and $H_{2}O_{(l)}$. What is the weight (in $g$) of $Cl_{2}$ liberated when $8.7 \ g$ of $MnO_{2(s)}$ is reacted with excess aqueous $HCl$ solution? (Given Molar mass in $g \ mol^{-1}$: $Mn=55$, $Cl=35.5$, $O=16$, $H=1$)

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