The average human body contains 6.20 L of blood with a concentration of 2.40×10−5 M . If a person ingests 8.00 mL of 13.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?
According to the equation:
Fe2+ + 6CN- ->[Fe(CN)6]4-
kf = 4.21*10^45
I calculated mols Fe^2+ as 6.2L x 2.4E-5 = 1.49E-4
CN^- 0.008L x 0.013 M = 1.04E-4
With such a large Kf I assumed equilibrium conditions would be FAR FAR to the right.
..........Fe^2+ + 6CN^- ==> [Fe(CN(6]^-4
initial..1.49E-4 mols
add..............1.04E-4mols
change.1.73E-5...1.04E-4.....1.73E-5
equil...?
% iron complexed = (1.73E-5/1.49E-4)*100 = 11+ %?
Thank you! I got it.
sex
Well, well, well, looks like someone's trying to get all scientific on me! But hey, I clown around with all kinds of questions, so here we go!
Let's break this down, my friend. We're trying to find the percentage of iron(II) in the blood that would be sequestered by the cyanide ion. To do that, we need to figure out how many moles of iron(II) are present in the blood and compare it to how many moles of cyanide would react with it. Alright, let's get our calculators ready!
First, we need to find the moles of NaCN ingested. We can use the formula:
moles = concentration x volume
So, moles of NaCN = (13.0 mM / 1000) x 8.00 mL = 0.104 moles
Now, let's calculate the moles of iron(II) in the blood. We know the concentration is 2.40×10−5 M and the volume is 6.20 L. It's time for some math magic:
moles of iron(II) = (2.40×10−5 M) x (6.20 L) = 1.488×10−4 moles
According to the equation, one mole of iron(II) reacts with six moles of cyanide. So, we multiply the moles of iron(II) by 6:
moles of cyanide needed = 1.488×10−4 moles x 6 = 8.928×10−4 moles
Now, to find the percentage of iron(II) in the blood that would be sequestered by the cyanide ion, we divide the moles of cyanide needed by the moles of iron(II) in the blood and multiply by 100:
percentage = (8.928×10−4 moles / 1.488×10−4 moles) x 100 ≈ 599.6%
Wowzers! Looks like we have a bit of a sequestration party going on here. According to my calculations, the percentage of iron(II) in the blood that would be sequestered by the cyanide ion is approximately 599.6%. That's one way to turn a clown's frown upside down!
Disclaimer: Please remember, my friend, that these calculations are just for fun and should not be taken as absolute truth. Always consult with a qualified professional for serious matters.
Now, who's up for some circus tricks?
To calculate the percentage of iron(II) in the blood that would be sequestered by the cyanide ion, we need to determine the number of moles of iron(II) and the number of moles of cyanide ion.
First, let's calculate the number of moles of cyanide ion (CN-) in 8.00 mL of 13.0 mM NaCN. We can use the formula:
Molarity (M) = moles of solute / volume of solution in liters
Molarity (mM) = moles of solute / volume of solution in milliliters
Converting 8.00 mL to liters:
8.00 mL = 8.00 * 10^(-3) L
Now we can calculate the moles of cyanide ion (CN-):
13.0 mM NaCN = 13.0 * 10^(-3) moles / 8.00 * 10^(-3) L
= 1.625 moles
Next, we need to calculate the number of moles of iron(II) in the blood. We can use the concentration of iron(II) in the blood and the volume of blood.
Converting 6.20 L to milliliters:
6.20 L = 6.20 * 10^3 mL
Now we can calculate the moles of iron(II):
2.40 * 10^(-5) M * 6.20 * 10^3 mL
= 0.149 moles
Now we can determine the number of moles of iron(II) sequestered by the cyanide ion. Looking at the balanced equation, we see that it takes 6 moles of cyanide ions (CN-) to sequester 1 mole of iron(II) (Fe2+).
Therefore, the number of moles of iron(II) sequestered = 6 * (1.625 moles of cyanide ion)
= 9.75 moles
Finally, to calculate the percentage, we divide the number of moles of iron(II) sequestered by the total moles of iron(II) in the blood and multiply by 100:
Percentage = (9.75 moles / 0.149 moles) * 100
= 6536.9%
Therefore, approximately 6,536.9% of iron(II) in the blood would be sequestered by the cyanide ion.