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$2K_3Fe(CN)_6+2KOH\rightarrow x+y+z$,$H_2O_2+z\rightarrow y+w$.The $y$ is

$\begin{array}{1 1}(a)\;H_2\\(b)\;2H_2O\\(c)\;H_2O\\(d)\;2HO\end{array}$

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1 Answer

$2K_3Fe(CN)_6+2KOH\rightarrow 2K_4Fe(CN)_6+H_2O+O$
$H_2O_2+O\rightarrow H_2O+O_2$
Hence $y$ is $H_2O$
Hence (c) is the correct answer.
answered Mar 25, 2014 by balaji.thirumalai
 
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