The average human body contains 6.50 L of blood with a Fe2+ concentration of 2.40×10−5M ....
The average human body contains 6.50 L of blood with a Fe2+ concentration of 2.40×10−5M . If a person ingests 5.00 mL of 23.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?
Homework Answers
Fe2+ + 6CN- = [Fe(CN)6]4-
Ratio 1:6
2.40×10−5 mol/L x 6.5 L = 1.56x10−4 mol Fe2+ in 6.50 L of blood
0.005 L x 0.023 M NaCN = 1.15x10-4 mol CN- ingested
The percentage of iron(II) in the blood sequestered by the cyanide ion:
100 x (1.15x10-4 / 6) / 1.56x10−4 = 12.3 %
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