The average human body contains 5.70 L of blood with a Fe2+ concentration of 3.20×10−5 M...

The average human body contains 5.70 L of blood with a Fe2+ concentration of 3.20×10−5 M . If a person ingests 7.00 mL of 24.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Expert Solution

Number of moles of NaCN , n = Molarity x volume in L

= 24.0 mM x 7.00mL

= 24.0x10^-3 x 7.00x10^-3 L

= 1.68x10^-4 mol

The balanced reaction between Fe2+ & NaCN is Fe(OH)₂ + 2 NaCN Fe(CN)₂ + 2 NaOH

From the above reaction

2 moles of NaCN reacts with 1 mole of Fe²⁺

1.68x10^-4 mol of NaCN reacts with (1.68x10^-4 )/2 = 8.4x10^-5 mole of Fe²⁺

So Concentration of Fe²⁺ that would be reacted with Cyanide ion is

= number of moles of Fe²⁺ reacted / total volume

= 8.4x10^-5 mol / 5.70 L
= 1.47x10^-5 M

Therefore the percentage of Iron(II) in the blood that sequestered is = [(1.47x10^-5 )/(3.20x10^-5)]x100

= 46.05 %

queen_honey_blossom answered 2 hours ago

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