How many moles of H_2O (per mole of Cl_2 evol | vedclass

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(D) The balanced chemical equation for the reaction between $KMnO_4$ and $HCl$ is:$2 KMnO_4 + 16 HCl \rightarrow 2 MnCl_2 + 5 Cl_2 + 2 KCl + 8 H_2O$Di

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$8/5$

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(D) The balanced chemical equation for the reaction between $KMnO_4$ and $HCl$ is:$2 KMnO_4 + 16 HCl ightarrow 2 MnCl_2 + 5 Cl_2 + 2 KCl + 8 H_2O$Dividing the entire equation by $2$ to normalize per mole of $KMnO_4$:$KMnO_4 + 8 HCl ightarrow MnCl_2 + \frac{5}{2} Cl_2 + KCl + 4 H_2O$From the stoichiometry,we see that $\frac{5}{2}$ moles of $Cl_2$ are produced along with $4$ moles of $H_2O$.To find the moles of $H_2O$ per mole of $Cl_2$,we calculate:$	ext{Ratio} = \frac{4 	ext{ moles } H_2O}{5/2 	ext{ moles } Cl_2} = 4 	imes \frac{2}{5} = \frac{8}{5}$ moles of $H_2O$.

How many moles of $H_2O$ (per mole of $Cl_2$ evolved) will be formed when $KMnO_4$ is added to $HCl$ solution?

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