Question

The average human body contains 5.30L of blood with a Fe2+ concentration of 2.30×10−5M . If a person ingests 11.0mL of 15.0mM NaCN , what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

I would look at it like this.
iron(II) + 6CN^- ==>Fe(CN)6^-4

So we have 5.30 L x 2.30 x 10^-5 M = ?? moles Fe(II).

We have 0.011 x 0.015 M NaCN = ?? moles CN^-.
Convert moles CN^- to moles iron(II). That equals the moles of iron(II) sequestered by CN^- so that number divided by iron initially x 100 is % iron sequestered.
Check my thinking.

Answer 2

I converted CN^- to moles iron(II)which is 2.75*10^-5, and initially 1022*10^-4 mols of Fe(II) were present. now i don't get the next step ..

Answer 3

correction: initially 1.22*10^-4 mols of Fe(II) were present

Answer 4

2.75 x 10^-5 moles Fe^+2 complexed.
You had 1.22 moles Fe^+ initially.
So % sequestered is
(2.75 x 10^-5/1.22 x 10^-4)*100 = ??

Answer 5

this still gives me the wrong answer =(
plz help..

Answer 6

22.6% sequestered should be the correct answer IF the complex is Fe(CN)6^-4. I don't know that it is. You might try Fe(CN)4^-2 and rework it that way. Before the first reponse, I look up to see how Fe is present in the blood. There is a mixture of iron(II) and iron(III). The iron(III) I know forms the Fe(CN)6^-3 and my best educated guess is that iron(II) also forms with the 6 addends. And the way the problem is stated, it appears to be neglecting any iron(III) there anyway. My point is that the iron(III) will comp;lex with the CN also but the problem doesn't tell us how much iron(III) is there; that's why I neglected it.