$KMnO_4 + HCl \rightarrow MnCl_2 + Cl_2 + KCl + H_2O$. In the above reaction,how many moles of $H_2O$ would be formed for each mole of $Cl_2$ liberated?
- A$\frac{5}{4}$
- B$\frac{4}{5}$
- C$\frac{5}{8}$
- D$\frac{8}{5}$
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