Question

The average human body contains 5.60 L of blood with a Fe2 concentration of 3.00×10−5 M . If a person ingests 9.00 mL of 11.0 mM NaCN,
what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Answer:

9.82% of iron (II) will be sequestered by cyanide

Explanation:

We should first consider that Iron (II) and cyanide react to form the following structure:

[Fe(CN)₆]⁻⁴

Having considered this:

5.60 Lt Fe(II) 3.00x10⁻⁵ M ,this is, we have 5.60x3x10⁻⁵ =  1.68x10⁻⁴ moles of Fe⁺² (in 5.60 Lt)

Then , we have 9 ml NaCN 11.0 mM:

9 ml = 0.009 Lt

11.0 mM (milimolar) = 0.011 M (mol/lt)

So: 0.009x0.011 = 9.9x10⁻⁵ moles of CN⁻ ingested

As we now that the complex structure is formed by 1 Fe⁺² : 6 CN⁻ :

9.9x10⁻⁵ moles of CN⁻ will use 1.65x10⁻⁵ moles of Fe⁺² (this is, this amount of iron (II) will be sequestered

[(1.65x10⁻⁵ sequestred Fe⁺²)/(1.68x10⁻⁴ total available Fe⁺²)x100

% sequestered iron (II) = 9.82%