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The average human body contains 5.60 L of blood with a Fe2 concentration of 3.00Ɨ10āˆ’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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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)ā‚†]ā»ā“

Having considered this:

5.60 Lt Fe(II) 3.00x10ā»āµ M ,this is, we have 5.60x3x10ā»āµ = Ā 1.68x10ā»ā“ moles of FeāŗĀ² (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 = 9.9x10ā»āµ moles of CNā» ingested

As we now that the complex structure is formed by 1 FeāŗĀ² : 6 CNā» :

9.9x10ā»āµ moles of CNā» will use 1.65x10ā»āµ moles of FeāŗĀ² (this is, this amount of iron (II) will be sequestered

[(1.65x10ā»āµ sequestred FeāŗĀ²)/(1.68x10ā»ā“ total available FeāŗĀ²)x100

% sequestered iron (II) = 9.82%

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