Determination of Iron by Reaction with Permanganate

Determination of Iron by Reaction With Permanganate [Document Subtitle] Brianna A., Mary, Maabo, & Zarel causation: The purpose of this lab was to determine the mass percent of conjure in an enigmatical mingled. Procedure: 1. Weigh out 1g of the unknown with a scale and pour into a clean Erlenmeyer flask. 2. gear up the buret. Get 50 mL of .05 M KMnO4 and rinse the buret with a couple mL. 3. Add 50 mL of 1 M H2SO4 to the consume in the Erlenmeyer flask. 4. Titrate the solution with the KMnO4 solution. If a light yellow color develops, add 3mL of 85% H3PO4. somersaulting on titrating until you get the first pink color that lasts for bimestrial than 15 to 30 seconds. 5. Repeat 2x more. Background: Oxidation is the way out of an electron darn reduction is gaining an electron. An oxidizing agent accepts electrons from another reactant while a reducing agent gives up electrons in night club to strangle the oxidation state of one of its atoms. Results: a. Data | pattern 1| precedent 2| Sample 3| stool of iron compound| .94g| .96g| .97g| Initial burette drill | 0 mL| 0 mL| 0 mL| concluding burette reading| 17 mL| 15 mL| 14.5 mL| muckle KMnO4 required| 17 mL| 15 mL| 14.5 mL| b. Calculations 1. MnO4- + 5Fe2+ +8H+ ? 4H2O +Mn2+ + 5Fe3 2.

| Sample 1| Sample 2| Sample 3| Moles KMnO4 required| 8.5 x 10-4 gram groineecule| 7.5 x 10-4 bulwark| 7.25 x 10-4 mol| Moles of Fe2+ in assay| 4.25 x 10-3 mol| 3.75 x 10-3 mol| 3.63x 10-3 mol| Mass of Fe in sample| .237 g| .209g| .202g| Mass of sample | .94g| .96g| .97g| % of Fe in sample| 25.! 2%| 21.8%| 20.8%| Sample 1: .017 L x .050 mol KMnO41L = .00085 mol MnO4 .00085 mol MnO4 x 5Fe2+1MnO4 = 4.25 x 10 -3 mol Fe2+ 4.25 x 10 -3 mol Fe2+ x 55.8g1 mol = .237g %= .237g.94 g x100= 25.2% Sample 2: .015 L x .050 mol KMnO41L = .00075 mol MnO4 .00075 mol MnO4 x 5Fe2+1MnO4 = .00375 mol Fe2+ .00375 mol Fe2+ x 55.8g1 mol = .209 g %=...If you desire to get a full(a) essay, order it on our website: BestEssayCheap.com

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