Question

# The average human body contains 6.50 L of blood with a Fe2+ concentration of 2.30×10−5 M...

The average human body contains 6.50 L of blood with a Fe2+ concentration of 2.30×10−5 M . If a person ingests 12.0 mL of 23.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Number of moles of Fe2+ , n = Molarity x volume in L

= 2.30x10-5 M x 6.50 L

= 1.495x10-4 moles

number of moles of NaCN = molarity x volume in L

= 23.0mM x 12.0 mL

= 23.0x10-3 M x 12.0x10-3 L

= 276x10-6 mol

Fe2+ + 6 CN- -----> [Fe(CN)6]4+

1 mole of Fe2+ reacts with 6 moles of CN-

M moles of  Fe2+ reacts with 276x10-6 mol of CN-

M = (1x276x10-6 )/ 6

= 4.6x10-5 moles

So percentage of Fe2+ is = [(4.6x10-5  mol)/ (1.495x10-4 mol) ]x100

= 30.8 %

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