SHOW SOLUTION A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(I1) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe2O3 (MM 159.69). 5Fe+2 + MnO4 - + 8H+ → 5Fe+3 + Mn +2 + 4H2O Answer: 32.08% Fe, 45. 86% Fe2O3

SHOW SOLUTION A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(I1) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe2O3 (MM 159.69). 5Fe+2 + MnO4 - + 8H+ → 5Fe+3 + Mn +2 + 4H2O Answer: 32.08% Fe, 45. 86% Fe2O3

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.34QAP

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A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(II) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe2O3 (Molar mass= 159.69). 5Fe+2 + MnO4- +8H+ → 5Fe+3 + Mn+2 + 4H2O
1. A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(II) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe₂O3 (MM 159.69). 5Fe +2 + MnO4 +8H+ → 5Fe+³+ Mn+² + 4H₂O
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