Answer:
The percentage is % [tex]Fe^{2+[/tex] [tex]= 57.3[/tex]%
Explanation:
From the question we are told that
The volume of blood in the human body is [tex]V = 5.0 0 \ L[/tex]
The concentration of [tex]Fe^{2+[/tex] is [tex]C_{F} = 1.10 *10^{-5} \ M[/tex]
The volume of NaCN ingested is [tex]V_N = 9.00 \ mL = 9.00 *10^{-3} \ L[/tex]
The concentration of NaCN ingested is [tex]C_N = 21.0 \ mM = 21.0 *10^{-3} \ M[/tex]
The number of moles of [tex]Fe^{2+[/tex] in the blood is
[tex]N_F = C_F * V[/tex]
substituting values
[tex]N_F = 1.10 *10^{-5} * 5[/tex]
[tex]N_F = 5.5*10^{-5} \ mols[/tex]
The number of moles of [tex]CN^{-}[/tex] ingested is mathematically evaluated as
[tex]N_C = C_N * V_N[/tex]
substituting values
[tex]N_C = 21*10^{-3} * 9 *10^{-3}[/tex]
[tex]N_C = 1.89 *10^{-4} \ mols[/tex]
The balanced chemical equation for the reaction between [tex]Fe^{2+[/tex] and [tex]CN^{-}[/tex] is represented as
[tex]Fe^{2+} + 6 CN^{-} \to [Fe(CN)_6]^{2-}[/tex]
From this reaction we see that
1 mole of [tex]Fe^{2+[/tex] will react with 6 moles of [tex]CN^{-}[/tex]
=> x moles of [tex]Fe^{2+[/tex] will react with [tex]1.89 *10^{-4} \ moles[/tex] of [tex]CN^{-}[/tex]
Thus
[tex]x = \frac{1.89 *10^{-4} * 1}{6}[/tex]
[tex]x = 3.15 *10^{-5}[/tex]
Hence the percentage of [tex]Fe^{2+[/tex] that reacted is mathematically evaluated as
% [tex]Fe^{2+[/tex] [tex]= \frac{3.15 *10^{-5}}{5.5*10^{-5}} * 100[/tex]
% [tex]Fe^{2+[/tex] [tex]= 57.3[/tex]%
