How much anhydrous stannous chloride (189.61 g/mol) is needed to completely reduce 1.050 g iron ore that has 26.0 % Fe (III)? [Given: FWFe = 55.845 g/mol] Half-reactions: Sn2+ --> Sn4+ + 2e- Fe2+ --> Fe3+ + e- A. 0.212 g B. 0.464 g

How much anhydrous stannous chloride (189.61 g/mol) is needed to completely reduce 1.050 g iron ore that has 26.0 % Fe (III)? [Given: FWFe = 55.845 g/mol] Half-reactions: Sn2+ --> Sn4+ + 2e- Fe2+ --> Fe3+ + e- A. 0.212 g B. 0.464 g

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 24Q: What is wrong with the following statement: The best concentration cell will consist of the...

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Question

How much anhydrous stannous chloride (189.61 g/mol) is needed to completely reduce 1.050 g iron ore that has 26.0 % Fe (III)? [Given: FWFe = 55.845 g/mol]
Half-reactions: Sn2+ --> Sn4+ + 2e-
Fe2+ --> Fe3+ + e-
A. 0.212 g

B. 0.464 g

C. 0.927 g

D.1.78 g

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