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?
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.
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 ..
correction: initially 1.22*10^-4 mols of Fe(II) were present
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 = ??
this still gives me the wrong answer =(
plz help..
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.
To calculate the percentage of iron(II) in the blood that would be sequestered by the cyanide ion, we need to follow a few steps.
Step 1: Calculate the number of moles of iron(II) ions in the blood.
Given that the blood volume is 5.30L and the concentration of Fe2+ ions is 2.30×10^(-5) M, we can use the formula:
moles of Fe2+ = concentration of Fe2+ × volume of blood
moles of Fe2+ = 2.30×10^(-5) M × 5.30L
Step 2: Calculate the number of moles of NaCN ingested.
Given that 11.0mL of a 15.0mM (millimolar) NaCN solution is ingested, we can use the formula:
moles of NaCN = concentration of NaCN × volume of NaCN
moles of NaCN = 15.0mM × 11.0mL
Step 3: Determine the stoichiometry between Fe2+ and CN- ions.
Referencing the balanced equation between Fe2+ and CN- ions, we can determine the stoichiometric ratio of Fe2+ to CN-, which is 1:2. This means that for every 1 mole of Fe2+, 2 moles of CN- are required.
Step 4: Calculate the maximum number of moles of Fe2+ that can be sequestered by the CN- ions.
Since the stoichiometric ratio is 1:2, the maximum number of moles of Fe2+ sequestered is half the number of moles of CN- ions ingested.
moles of Fe2+ sequestered = (1/2) × moles of NaCN
Step 5: Calculate the percentage of Fe2+ sequestered.
Finally, we can calculate the percentage of Fe2+ that would be sequestered by the cyanide ion by dividing the moles of Fe2+ sequestered by the moles of Fe2+ in the blood and multiplying by 100.
percentage of Fe2+ sequestered = (moles of Fe2+ sequestered / moles of Fe2+ in blood) × 100
By following these steps, you can find the desired percentage of iron(II) in the blood that would be sequestered by the cyanide ion.