Question

The average human body contains 5.90 L of blood with a Fe2+ concentration of 3.10×10−5 M . If a person ingests 7.00 mL of 18.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Moles of Fe⁺² = Molarity X volume = 3.1 X 10^-5 X 5.9 = 1.829 X 10^-4

Moles of CN- = 18 X 10^-3 X 7 X 10^-3 moles = 1.26 X 10^-4 moles

The reaction of Ferrous ion and CN-

        Fe⁺²   + 6CN- ---> [Fe(CN)₆]^-2

So one mole of Fe+2 reacts with 6 moles of CN-

Therefore 1 mole of CN- will react with 1/6 moles of Fe+2

1.26 X 10^-4 moles of CN- will react with = 1.26 X 10^-4 / 6 moles of Fe⁺² = 0.21 X 10^-4

Percentage of Fe+2 reacted = 0.21 X 10^-4 X 100 / 1.829 X 10^-4 = 11.18%