A 2.16 g sample containing both Fe and V was dissolved under certain conditions

Question

A 2.16 g sample containing both Fe and V was dissolved under certain conditions and diluted to 500.00 mL. A first 50.00 taken and passed through a Walden reductor to form Fe^2+ and VO^2+ ions. The titration of this solution required 16.84 mL of 0.1000 M Ce^4+ to reach end point. A second 50.00 mL aliquot was passed through a Jones reductor to form Fe^2+ and V^2+ ions. The titration of the second solution required 44.26 mL of 0.1000 M Ce^4+ solution to reach an end point. Calculate the percentage of Fe and V in the sample.

Explanation / Answer

The Walden reductor reduces Fe(III) to Fe(II) and the Jones reductor V(V) to V(II). It also seems that Ce4+ reduces to Ce3+. This would lead to believe that in the iron case you have a 1:1 reaction and in the vanadium case a 1:3 reaction.

For the Iron in the 50ml aliquot there is the same number of moles of iron as the number of moles of Ce titrated.

Number of moles of Ce= Number of moles of Fe = Concentration*Volume= 0.1*16.84=1.684 moles

Now, mass of iron = number of moles* atomic mass of iron =1.684*55.85= 94.0514gm=0.0940514Kg

Now %=(mass/2.559)*100=(0.094051/2.559)*100=3.68% iron

For the Vanadium in the 50ml aliquot, there is 1/3 the number of moles of Ce4+ titrated.

Number of moles of Ce= 1/3Number of moles of V = 1/3Concentration*Volume= 1/30.1*16.84=0.561 moles

Now, mass of iron = number of moles* atomic mass of iron =0.561*50.94= 28.59gm=0.02859Kg

Now %=(mass/2.559)*100=(0.02859/2.559)*100=1.12% iron

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