The average human body contains 6.00 L of blood with a Fe2+ concentration of 2.00×10−5 M . If a person ingests 6.00 mL of 16.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?
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To find out the percentage of iron(II) in the blood that would be sequestered by the cyanide ion, we need to calculate the moles of iron(II) in the blood and the moles of iron(II) that would react with the cyanide ion.
First, we calculate the moles of iron(II) in the blood:
Moles of Fe2+ in blood = (concentration of Fe2+ in blood) x (volume of blood)
= (2.00×10^(-5) M) x (6.00 L)
Next, we calculate the moles of iron(II) that would react with the cyanide ion:
Moles of Fe2+ reacting with cyanide = (volume of NaCN) x (concentration of NaCN)
= (6.00 mL) x (16.0 mM)
Now, let's convert the volume of blood and NaCN to the same unit (liters) for easier calculations:
Volume of blood = 6.00 L
Volume of NaCN = 6.00 mL = 6.00 x 10^(-3) L
Now, let's substitute the values and calculate:
Moles of Fe2+ in blood = (2.00×10^(-5) M) x (6.00 L) = 1.20 x 10^(-4) moles
Moles of Fe2+ reacting with cyanide = (6.00 x 10^(-3) L) x (16.0 x 10^(-3) M) = 9.60 x 10^(-5) moles
Finally, we can calculate the percentage of iron(II) sequestered by the cyanide ion:
Percentage sequestered = (moles of Fe2+ reacting with cyanide / moles of Fe2+ in blood) x 100
Percentage sequestered = (9.60 x 10^(-5) moles / 1.20 x 10^(-4)) x 100
Percentage sequestered = 80.0%
Therefore, approximately 80.0% of the iron(II) in the blood would be sequestered by the cyanide ion.