Question

The average human body contains 5.10 L of blood with a Fe2+ concentration of 2.60×10−5mol L−1. If a person ingests 12.0 mL of 18.0 mol L−1 NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Express the percentage numerically.

Answer 1

given data's volume of blood = solo feta concentration = 2.60x105 mol concentration of MacM = 18.0 mol MI volume ingests of MaCM = 1200 me Reaction between reta and MacM Fe2+ + 2Nach PelCN)2 + anat moles of fe2+ = (concentration of fe 2+ ) x ( volume of blood) = (2-60x10s mol ) x (5:10 L) = 1:326x10-4 mod fe24 moles of MaCM = (concentration of MaCM) xe volume) = (18.0 mol/ ( 12418-32) = 0.216 mol MOCH Feet + 2Nach Fe(CM/2+2 Mat By mole to mole ratio 1 mol feat 0.108 mol Felt reacted 8.216 mol Hach X Scanned with CamScanner 2- мо| NaCH

% of tron(11) = mote of Fe2t reathed mole of fezt = 0.108 mol 1:326410-4 mol = 814.47 some error in concentration of Mach there, that's why auwes is going is beyond 100%. Scanned with CamScanner