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

$\begin{array}{1 1}(a)\;K_4Fe_4(CN)_2\\(b)\;2K_4Fe(CN)_6\\(c)\;2K_4Fe_2(CN)_6\\(d)\;2K_4Fe_4(CN)_3\end{array}$

$2K_3Fe(CN)_6+2KOH\rightarrow 2K_4Fe(CN)_6+H_2O+O$
Hence $x$ is $2K_4Fe(CN)_6$
Hence (b) is the correct answer.
answered Jan 24, 2014