The average human body contains 6.30 L of blood with a Fe^2+ concentration of 2

Question

The average human body contains 6.30 L of blood with a Fe^2+ concentration of 2 30x10^-5 M. If a person ingests 5.00 mL of 12.0mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion? The reaction between iron in the blood and the cyanide ion is. Fe^2+ + 6CN- arrow [Fe(CN)6]^4- (Remember to convert mM to M) a Calculate the number of moles of each reactant, Fe^2+ and CN^- using the given volumes and concentrations moles Fe^2+ = moles CN- = b. Use stoichiometry to calculate the number of moles of Fe^2+ that react. Assume the reaction goes to completion. moles react c. Express the amount of Fe^2+ that reacts as a percentage of the original amount in the blood

Explanation / Answer

I will answer only the first paragraph a), if you want the answers to the other questions attach them to a new question.

a)

Fe mol=2.3*10-5mol/L*6.3L=1.45*10-4

CNfmol=0.53

CN mol=(0.53*0.012*0.005)=3.18*10-5

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