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 mL aliquot was 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

First 50 ml aliquot

moles of Ce4+ used = 0.1 M x 16.84 ml = 1.684 mmol

moles of Fe2+ present = 1.684 mmol

moles of Fe2+ in original 500 ml solution = 16.84 mmol

mass of Fe in sample = 0.01684 x 55.845 = 0.94 g

percent Fe in the sample = 0.94 x 100/2.16 = 43.52%

Second 50 ml aliquot

moles of Ce4+ used = 0.1 M x 44.26 ml = 4.426 mmol

moles of V2+ present = 4.426/3 = 1.475 mmol

moles of V in original 500 ml solution = 14.75 mmol

mass of V in sample = 0.01475 x 50.94 = 0.751 g

percent V in sample = 0.751 x 100/2.16 = 34.78%

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