Question

The average human body contains 5.50 L of blood with a Fe2+ concentration of 2.20×10−5 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 –Given, [NaCN] = 21.0 mM , volume = 19.0 mL

[Fe²⁺] = 2.20*10^-5 M , volume = 5.50 L

Reaction between Fe²⁺ and CN^-

Fe²⁺ + 6 CN^- -----> [Fe(CN)₆]^4-

Now we need to calculate the moles of Fe²⁺

Moles of Fe²⁺ = 2.20*10^-5 M * 5.50 L

                       = 1.21*10^-4 moles

We know ,

1 mM = 0.001 M

So, 21.0 mM = ?

= 0.021 M

Moles of CN^- = 0.021 M * 0.009 L

                     = 1.89*10^-4 moles

Moles of Fe²⁺ used in the reaction -

6 moles of CN^- = 1 moles of Fe²⁺

So, 1.89*10^-4 moles CN^- = ?

= 3.15*10^-5 moles of Fe²⁺

Moles of Fe²⁺ remaining = 1.21*10^-4 moles - 3.15*10^-5 moles

                                         = 8.5*10^-5 moles of Fe²⁺

percent of the iron(II) in the blood would be sequestered by the cyanide ion

percent of the iron(II) = 8.5*10^-5 moles of Fe²⁺ / 1.21*10^-4 moles * 100 %

                                    = 74.0 %