Question

The average human body contains 6.50 L of blood with a Fe2+ concentration of 2.30×10−5 M . If a person ingests 12.0 mL of 23.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Number of moles of Fe2+ , n = Molarity x volume in L

= 2.30x10^-5 M x 6.50 L

= 1.495x10^-4 moles

number of moles of NaCN = molarity x volume in L

= 23.0mM x 12.0 mL

= 23.0x10^-3 M x 12.0x10^-3 L

= 276x10^-6 mol

Fe²⁺ + 6 CN^- -----> [Fe(CN)₆]⁴⁺

1 mole of Fe²⁺ reacts with 6 moles of CN^-

M moles of  Fe²⁺ reacts with 276x10^-6 mol of CN-

M = (1x276x10^-6 )/ 6

= 4.6x10^-5 moles

So percentage of Fe2+ is = [(4.6x10^-5  mol)/ (1.495x10^-4 mol) ]x100

= 30.8 %