Question

The average human body contains 6.40 L of blood with a Fe2+ concentration of 1.40×10⁻⁵M . If a person ingests 11.0 mL of 19.0 mMNaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Express the percentage numerically.

Answer 1

Molarity = Moles/Liter

Moles Fe²⁺ = (6.40 L) x (1.40 x 10^-5 M) = 8.96 x 10^-5

Moles CN^- = (0.011 L) x (1.9 x 10^-3 M) = 2.09 x 10^-4

Now,

The reaction that occurs is :

Fe²⁺ + 6 CN^- = [Fe(CN)₆]^4-

Hence, the ratio is 1 :6

moles Fe²⁺ that react with CN^- = (2.09 x 10^-4) / 6

                                                 = 3.48 x 10^-5

% Fe2+ that reacts = [ (3.48 x 10^-5) / (8.96 x 10^-5) ] x 100

                              = 38.84 %