[sarah]

A 2.559g of sample containing both Fe and V was dissolved under conditions that converted the elements Fe(III) and V(V). The solution was diluted to a 500.0mL, and a 50.00mL aliquot just passed through a Walden reductor and titrated with 17.74mL o 0.1000M Ce4+. A second 50.00mL aliquot was passed through a Jones reductor and required 44.67mL of the same Ce4+solution to reach an end point. Calculate the percentage of Fe2O3 and V2O5 in the sample

[Sam (NG)]

I'm not used to doing this, so you'll need to get confirmation from someone else and not take my word for it (i'm not an analytical chemist).  From what i just read i figure that 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 me 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.

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

Multiply these by 10 and you have the number of moles of Fe(III) and V(V) in the original sample.  These can then be converted to masses and divided by 2.559 (and multiplied by 100) to give the percentage in the sample.

[Borek]

sarah: please read forum rules.