In the reaction,$2KMnO_4 + 16HCl \rightarrow 5Cl_2 + 2MnCl_2 + 2KCl + 8H_2O$,the reduction product is

  • A
    $Cl_2$
  • B
    $MnCl_2$
  • C
    $H_2O$
  • D
    $KCl$

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