Question

The average human body contains 6.10 L of blood with a Fe2+ concentration of 1.50×10−5 M . If a person ingests 12.0 mL of 14.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Answer: The percentage of iron(II) in the blood sequestered by the cyanide ion is 18.4%

Explanation:

Having the reaction NaCN(ac) + Fe2+ ⇒ CNFe + Na+ ; you could follow this procedure:

1. Find the number of moles of Fe2+ in the 6.10L of blood and the number of moles of CN- in the 0.012L (0.012L=12mL) solution.

First, notice that the amount of CN- moles in the 12mL solution is the same as if we were looking for the amount of moles NaCN in the solution. This is because most sodium salts dissociate completely in aqueous solution.

Then having the Molar concentrations of both CN- and Fe2+:

Fe2+ concentration = 1.5x$10^{-5}$ M = 1.5x$10^{-5}$moles (n)/L

Moles de Fe2+ en 6.01L= (1.5x$10^{-5}$moles (n)/L)*6.01L= 9.15x$10^{-5}$ moles

NaCN or CN- concentration = 14mM=0,0014M=(0,0014n/L)

Moles de CN- en 0,012L= (0,0014n/L)*0,012L= 1.68x$10^{-5}$ moles

2. Understand how many moles of Fe2+ are sequestered by CN- :

Beacause 1 mol of CN- react (sequester) with 1 mol of Fe2+ to form 1 mol of CNFe, then 1.68x$10^{-5}$ moles of CN- is going to sequester 1.68x$10^{-5}$ moles of Fe2+ to form 1.68x$10^{-5}$ moles of CNFe.

3. Calculate the percentage of Fe2+ sequestered:

% Fe2+ = (moles of Fe2+ sequestered/Total moles of Fe2+ in the blood)*100

⇒ % Fe2+ = (1.68x$10^{-5}$) / (9.15x$10^{-5}$)=0.184*100 = 18.4% Percentage of Fe2+ moles sequestered by cyanide ion (CN-).