Question

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?

Answer 1

Answer:

The  percentage is    % $Fe^{2+$   $= 57.3$%

Explanation:

From the question we are told that  

       The volume of blood in the human body  is  $V = 5.0 0 \ L$

       The  concentration of  $Fe^{2+$ is  $C_{F} = 1.10 *10^{-5} \ M$

        The volume  of  NaCN  ingested is  $V_N = 9.00 \ mL = 9.00 *10^{-3} \ L$

       The concentration of  NaCN ingested is  $C_N = 21.0 \ mM = 21.0 *10^{-3} \ M$

The  number of moles of   $Fe^{2+$ in the blood is  

                    $N_F = C_F * V$

substituting values  

                    $N_F = 1.10 *10^{-5} * 5$

                    $N_F = 5.5*10^{-5} \ mols$

The  number of moles of  $CN^{-}$ ingested is  mathematically evaluated as

           $N_C = C_N * V_N$

substituting values    

          $N_C = 21*10^{-3} * 9 *10^{-3}$

          $N_C = 1.89 *10^{-4} \ mols$

The balanced chemical equation for the reaction  between   $Fe^{2+$ and   $CN^{-}$  is  represented as

          $Fe^{2+} + 6 CN^{-} \to [Fe(CN)_6]^{2-}$

From this  reaction we see that  

         1 mole  of    $Fe^{2+$  will react with 6  moles of  $CN^{-}$

=>         x  moles of  $Fe^{2+$ will react with   $1.89 *10^{-4} \ moles$ of  $CN^{-}$

Thus  

         $x = \frac{1.89 *10^{-4} * 1}{6}$

        $x = 3.15 *10^{-5}$

Hence the percentage  of  $Fe^{2+$  that reacted is  mathematically evaluated as

       

       %  $Fe^{2+$   $= \frac{3.15 *10^{-5}}{5.5*10^{-5}} * 100$

        %  $Fe^{2+$   $= 57.3$%