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##### Exactly 1.0065 g of an unknown solid containing Fe2(SO4)3

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Question

Exactly 1.0065 g of an unknown solid containing Fe2(SO4)3 and non-ionogenic impurities was dissolved in about 25 mL of water and placed on a column of Rexyn 101 ion exchange resin. The column was then eluted with 200 mL of water and the total eluent collected in a 250 mL volumetric flask and diluted to the mark with water. When a 50 mL aliquot of this solution was titrated, 11.35 mL of 0.10359 M NaOH solution was required to reach the end point. Calculate the percentage of Fe2(SO4)3 in the unknown solid.;This is the answer: 38.93% Fe2(SO4)3.;I need an explanation please!;Thanks:);Additional Requirements;Min Pages: 1;Level of Detail: Show all work;Other Requirements: I stuck on the last part I got the moles of the solution is 0.004703 moles.. if it was pure it is easy but how can i get the percentage of Fe2(SO4)3.

Paper#16156 | Written in 18-Jul-2015

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