Question

The average human body contains 5.80 L of blood with a Fe2+ concentration of 3.20×10−5 M . If a person ingests 11.0 mL of 22.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion? Express the percentage numerically.

Answer 1

moles of Fe2+ = 5.80 x 3.20 x 10^-5

                        = 1.856 x 10^-4 mol

moles of NaCN = 11.0 x 10^-3 x 22.0 x 10^-3

                          = 2.42 x 10^-4 mol

moles of CN- = 2.42 x 10^-4 mol

Fe2+ (aq)       +     6 CN- (aq)   -----------> [Fe(CN)6]4-

   1                     6                                   1

1.856 x 10^-4      2.42 x 10^-4

moles of Fe2+ sequesterd = 2.42 x 10^-4 / 6 = 4.033 x 10^-5

% = (4.033 x 10^-5 / 1.856 x 10^-4 ) x 100

   = 21.73 %