A 2.65 g sample containing both F c and V was dissolved under certain conditions

Question

A 2.65 g sample containing both F c and V was dissolved under certain conditions and diluted to 500.00 m L. A first 50.00 m L aliquot was taken and passed through a Waiden reductor to form Fe2+ and VO2* ions. The titration of this solution required 12.84 mL of 0.1000 M Cc4 to reach end point. A second 50.00 mL aliquot was passed through a Jones reductor to form Fe2* and V2' ions. The titration of the second solution required 40.26 mL of 0.1000 M Cc4 solution to reach an end point. Calculate the percentage of Fe and V in the sample.

Explanation / Answer

Let the mass of Fe be 'y' gram

Hence mass of V = (2.65 - y) g

Moles of Fe in the sample = mass / molecular mass = y g / 56.0 g/mol = y/56 mol

Moles of V in the sample = mass / molecular mass = (2.65 - y) g / 51.0 g/mol = (2.65 - y) / 51.0 mol

volume of the solution prepared, V = 500.00 mL

Case-1: Treatment with Walden reductor:

Volume of the aliquot taken = 50.00 mL (1/10th of 500 mL)

Moles of Fe in 50 mL =  y/56 mol /10 = y/560 mol

Moles of V in 50 mL =   (2.65 - y) / 51.0 mol / 10 = (2.65 - y) / 510 mol

Moles of Ce required = MxV = 0.10000 x 0.01284 L = 0.001284 mol

From the given balanced chemical reactions it is clear that

Moles of Ce = Moles of Fe + Moles of V

=>  0.001284 mol =  y/560 mol +  (2.65 - y) / 510 mol   

by solving the above equation we can find the value of y and hence the percentage

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.