Question

The average human body contains 5.60L of blood with a Fe2+ concentration of 3.20

Answer 1

  assuming that 1 CN- attaches to 1 Fe2+ and that "sequesters" the iron...

moles Fe attached to CN-.....
12.0 mL NaCN x (1 L / 1000 mL) x (23.0 mmole / L) x (1 mole / 1000 mMole) x (1 mole CN- / 1 mole NaCN) x (1 mole Fe2+ / 1 mole CN-) = 2.76x10^-4 moles Fe2+

total moles Fe2+ in your blood...
5.60 L x (3.20x10^-5 moles / L) = 1.792x10^-4 moles Fe2+

% Fe attached to CN- = 100%.. there is more CN- than Fe2+

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Now that I've read Joshua's answer, I guess it depends on how you intrepret this problem... if you mean this...

Fe2+ + 2 CN- ------> Fe(CN)2, then by all means...

moles Fe attached to CN- goes by this...
12.0 mL NaCN x (1 L / 1000 mL) x (23.0 mmole / L) x (1 mole / 1000 mMole) x (1 mole CN- / 1 mole NaCN) x (1 mole Fe2+ / 2 mole CN-) = 1.39x10^-4 moles Fe2+

and you end up with 1.39 / 1.792 x 100% = 77.57% bound Fe..

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however....

Fe2+ in blood is predominantly found in hemoglobin. The chemistry of hemoglobin binding with O2 and CN- is not a chemical reaction but rather a coordination complexation.

Each heme group contains 1 Fe2+ which can reversibly bind with 1 O2. It does NOT react with O2. Same goes for carbon monoxide and CN-. 1 CN- ion "binding" to that Fe2+, which it will because the Fe2+ / CN- complexation is preferred over Fe2+ / O2, will prevent the complexation of the heme Fe2+ with O2 thus "sequestering" the Fe2+

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so my advice to you is this..... submit either answer or both and explain your rationale then ask your instructor whether 1 CN- sequesters 1 Fe2+ or whether 2 CN- sequesters 1 Fe2+...

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