question_answer

Permanent hardness is due to \[C>B>A\] and \[F{{e}_{0.93}}{{O}_{1.00}}\] of \[2C{{u}_{2}}O+C{{u}_{2}}S\xrightarrow{{}}6Cu+S{{O}_{2}}\] and \[2C{{u}_{2}}S+3{{O}_{2}}\xrightarrow{{}}2C{{u}_{2}}O+S{{O}_{2}}\] and is removed by the addition of  \[2CuFe{{S}_{2}}+{{O}_{2}}\xrightarrow{{}}C{{u}_{2}}S+FeS+S{{O}_{2}}\] \[2FeS+3{{O}_{2}}\xrightarrow{{}}2FeO+2S{{O}_{2}}\] \[q=\Delta U\] If hardness is 100 ppm \[O\to D\], amount of\[O\to C\] required to soften 10 L of hard water, is

A)  \[O\to B\]                         

B)  \[O\to A\]

C)  \[A{{g}_{2}}O+{{H}_{2}}O+2{{e}^{-}}\xrightarrow{{}}2Ag+2O{{H}^{-}}\]               

D)  \[Fe(III)\]