A 3.994 g sample containing Fe, Ti, and other inert material was dissolved and d

Question

A 3.994 g sample containing Fe, Ti, and other inert material was dissolved and diluted to form 250.0 mL of solution. The dissolution conditions were such that Fe was converted to Fe(III) and Ti was converted to Ti(IV) in solution. A 25.00-mL aliquot of the solution was then passed through a Walden reductor and titrated with 0.2449 M Ce4 . The endpoint was reached following the addition of 20.08 mL of the Ce4 solution. A second 50.00-mL aliquot of the solution was passed through a Jones reductor and titrated with the same Ce4 solution. The endpoint was reached following the addition of 52.82 mL of the Ce4 solution. Calculate the mass percent of Fe and Ti in the sample.

Explanation / Answer

Amount of Fe (III) in 25mL aliquote : reaction in walden reductor

25 mL * M = 0.2449 M * 20.08 mL

or, M = 0.2449 M * 20.08 mL/ 25mL = 0.197 M

molarity of Fe = 0.197 M = 0.197moles/L

moles of Fe in 250 mL solution = 0.197 *250mL/1000mL = 0.049 moles

moles of Fe = 0.049 moles *56 g/mol = 2.74 gm

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Amount of Ti + Fe : Reaction in Jones reductor

Molarity of Ti + Fe in 50mL aliquote = 52.82 * 0.2449/50 = 0.26 M = 0.26 moles/L

amount of Ti + Fe in 250mL = 0.26 8250mL/1000mL = 0.065 moles

moles of Ti present in 250mL = 0.065 -moles of Fe = 0.065-0.049 = 0.016 moles

amount of Ti = 0.016 * 47.87g/mol = 0.76 gm

mass % of Fe = 2.74/3.99 *100 = 68.6 %

mass of Ti = 0.76 /3.994 *100 = 19.02 %

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