The possible formula for Prussian blue is ....

  • A
    $Fe_3[Fe(CN)_6]_2$
  • B
    $Fe_2[Fe(CN)_6]_3$
  • C
    $Fe_4[Fe(CN)_6]_3$
  • D
    $Fe_3[Fe(CN)_6]_4$

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What product is formed by mixing the solution of $K_4[Fe(CN)_6]$ with the solution of $FeCl_3$?

Assign $A, B, C, D$ from the given type of reaction.
$2K_3[Fe(CN)_6] + 3FeCl_2 \longrightarrow Fe_3[Fe(CN)_6]_2 \downarrow + 6KCl$

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