The goal of this exercise is to demonstrate your understanding of the material.

Question

The goal of this exercise is to demonstrate your understanding of the material. Therefore, you must show alt your work in sufficient detail so that I can follow your reasoning. An answer alone without all your work will receive no credit. Points will be deducted for lack of neatness. A sample is analyzed to determine its iron content (as Fe^2+) via a redox titration with potassium chromate as the titrant. In the titration Fe^2+ is oxidized to Fe^3- and CrO_4^2-: is reduced to Cr^3+. What is the percent by mass of iron in the sample if 0.9087 g of the sample required 45.68 mL of a 0.04322 M K_2CrO_4 solution to reach the endpoint?

Explanation / Answer

reaction for this titration is

6Fe2+ + Cr2O72¯ + 14H+ ---> 6 Fe3+ + 2Cr3+ + 7H2O

mole of dicromate =(0.04322mol/L)(0.04568L)=0.0019742896mol

A/C to reaction

0.0019742896mol x (6 mol Fe(II) / 1 mol dichromate)=0.011845737mol Fe^2+

0.011845737mol Fe^2+ x 55.845 g/mol =0.6615252g

Percent of iron in the sample: 0.6615252g /0.9087g x100 =72.79%

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