Chemistry
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?
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The percentage of iron(II) in the blood sequestered by the cyanide ion can be calculated by the following equation:
% of Fe2+ sequestered = [(11 mL x 15.0 mM)/ (5.3 L x 2.3 x 10-5 M)] x 100
% of Fe2+ sequestered = 0.033%
So, 0.033% of the iron(II) in the blood would be sequestered by the cyanide ion due to the ingestion of 11.0mL of 15.0mM NaCN.
