The average human body contains 5.00 L of blood with a Fe2+ concentration of 1.10×10−5 M . If a person ingests 9.00 mL of 21.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

The  percentage is    %   %

Explanation:

From the question we are told that  

       The volume of blood in the human body  is  

       The  concentration of   is  

        The volume  of  NaCN  ingested is  

       The concentration of  NaCN ingested is  

The  number of moles of   in the blood is  

substituting values  

The  number of moles of   ingested is  mathematically evaluated as

substituting values    

The balanced chemical equation for the reaction  between   and    is  represented as

From this  reaction we see that  

         1 mole  of      will react with 6  moles of  

=>         x  moles of   will react with   of  

Thus  

Hence the percentage  of    that reacted is  mathematically evaluated as

       %    

        %     %

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

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