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11 December, 23:53

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?

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  1. 12 December, 03:01
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    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%
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