The average human body contains 6.30 L of blood with a Fe2+ concentration of 1.60×10−5M ....

The average human body contains 6.30 L of blood with a Fe2+ concentration of 1.60×10⁻⁵M . If a person ingests 9.00 mL of 13.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Expert Solution

no of moles Fe^2+ = molarity * volume in L

= 1.6*10^-5 * 6.3 = 0.0001008moles

no of moles of CN^- = molarity * volume in L

= 0.013*0.009 = 0.000117moles

Fe^2+ + 6CN^- --------------> [Fe(CN)6]^4-

moles of Fe^2+ that react with CN^- = 0.000117/6 = 1.95*10^-5moles

%Fe^2+ = 0.0000195*100/0.0001008 = 19.34% >>>>>answer

queen_honey_blossom answered 2 years ago

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