Question

The average human body contains 5.40 L of blood with a Fe2+ concentration of 2.70×10−5 M . If a person ingests 11.0 mL of 25.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

We have 5.4 L of blood with 2.7 x 10^-5 mol/L concentration. We get the number of moles:

5.4 L * 2.7 x 10^-5 mol/L = 0.0001458 moles of Fe⁺²

The person ingests 0.011 L of 0.025 M NaCN, hence it will ingest:

0.011 L * 0.025 mol/L = 0.000275 moles of NaCN

The iron ion will react with NaCN as follows:

Fe⁺² + 2NaCN -> Fe(CN)₂ + 2Na⁺

So we can convert the number of moles that NaCN will consume of Fe⁺²:

0.000275 moles of NaCN * (1 mol of Fe⁺² / 2 moles of NaCN) = 0.0001375 moles of iron ion

Percentage sequestred: (0.0001375 / 0.0001458) * 100% = 94.3%