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 redactor 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

For Fe

moles of Ce4+ used for 50 ml aliquot = 0.1 M x 16.84 ml = 1.684 mmol

mass of Fe in 500 ml solution = 1.684 x 55.845/1000 = 0.094 g

mass of Fe in 500 ml solution = 0.094 x 10 = 0.94 g

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

For V

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

moles of V = 4.426 - 1.684 = 2.742 mmol

moles of V in original 500 ml solution = 2.742 x 10 = 27.42 mmol

mass V in sample = 27.42 mmol x 50.94 g/mol/1000 = 1.37 g

percent V in sample = 1.37 x 100/2.16 = 63.42%

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