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 Ce4+ 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. (5)

Explanation / Answer

moles of Ce4+ needed to reach first end point through walden reductor = 0.1 M x 16.84 ml

                                                                                                                  = 1.684 mmol

moles of Fe present = 1.684 mmol in 50 ml

mass of Fe present in 500 ml = 1.684 x 10 x 55.845/1000 = 0.940 g

% Fe in sample = 0.940 x 100/2.16 = 43.52%

moles of Ce4+ needed to reach second end point through jones reductor = 0.1 M x 44.26 ml

                                                                                                                     = 4.426 mmol

moles of V present = 4.426/3 mmol in 50 ml

mass of V present in 500 ml = 4.426 x 10 x 50.94/3 x 1000 = 0.751 g

% V in sample = 0.751 x 100/2.16 = 34.77%

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