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%
