Question

The average human body contains 6.40 L of blood with a Fe2+ concentration of 1.40×10−5M ....

The average human body contains 6.40 L of blood with a Fe2+ concentration of 1.40×10−5M . If a person ingests 11.0 mL of 19.0 mMNaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Express the percentage numerically.

Homework Answers

Answer #1

Molarity = Moles/Liter

Moles Fe2+ = (6.40 L) x (1.40 x 10-5 M) = 8.96 x 10-5

Moles CN- = (0.011 L) x (1.9 x 10-3 M) = 2.09 x 10-4

Now,

The reaction that occurs is :

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

Hence, the ratio is 1 :6

moles Fe2+ that react with CN- = (2.09 x 10-4) / 6

                                                 = 3.48 x 10-5

% Fe2+ that reacts = [ (3.48 x 10-5) / (8.96 x 10-5) ] x 100

                              = 38.84 %

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