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 <u>the amount of CN- moles in the 12mL solution is the same as if we were looking for the amount of moles NaCN</u> in the solution. This is because most sodium salts dissociate completely in aqueous solution.
Then having the Molar concentrations of both CN- and Fe2+:
<u><em> Fe2+ concentration =</em></u> 1.5x M = 1.5x
moles (n)/L
<u><em> Moles de Fe2+ en 6.01L=</em></u> (1.5xmoles (n)/L)*6.01L= 9.15x
moles
<em><u> NaCN or CN- concentration =</u></em> 14mM=0,0014M=(0,0014n/L)
<u><em> Moles de CN- en 0,012L=</em></u> (0,0014n/L)*0,012L= 1.68x 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 moles of <u><em>CN- is going to sequester 1.68x
moles of Fe2+</em></u> to form 1.68x
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) / (9.15x
)=0.184*100 = 18.4% Percentage of Fe2+ moles sequestered by cyanide ion (CN-).