Explanation The net balanced chemical reaction for the redox titration between sodium oxalate and M n O 4 − is as follows. 5 C 2 O 4 2 − ( a q ) + 2 M n O 4 − ( a q ) + 16 H + ( a q ) → 10 C O 2 ( g ) + 2 M n 2 + ( a q ) + 8 H 2 O ( l ) So, 5 moles of oxalate ( C 2 O 4 2 − ) requires 2 moles of M n O 4 − . Therefore, the mole ratio between sodium oxalate and MnO 4 − is 5:2. The molar mass of sodium oxalate is 133.99 g/mol. The mass of sodium oxalate is 0.339 g. Calculate the number of moles of sodium oxalate as follows: 2.53 × 10 − 3 m o l N a 2 C 2 O 4 × 2 m o l M n O 4 − 5 m o l N a 2 C 2 O 4 = 1.01 × 10 − 3 m o l M n O 4 − Calculate the number of moles of M n O 4 − as follows: 2.53 × 10 − 3 m o l N a 2 C 2 O 4 × 2 m o l M n O 4 − 5 m o l N a 2 C 2 O 4 = 1.01 × 10 − 3 m o l M n O 4 − Calculate the molarity of M n O 4 − solution as follows: 1.01 × 10 − 3 m o l M n O 4 − 50.1 m L × 1000 m L 1 L = 2

Principles Of Chemistry: A Molecular Approach, Loose-leaf Edition (4th Edition)

4th Edition

ISBN: 9780134989099

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 19, Problem 135E

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Interpretation: The percentage of uranium present in the given sample is to be calculated.

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Chapter 19, Problem 134E

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Principles Of Chemistry: A Molecular Approach, Loose-leaf Edition (4th Edition)

Ch. 19 - Prob. 1SAQ Ch. 19 - Q2. Which statement is true for voltaic cells? a)...Ch. 19 - Prob. 3SAQ Ch. 19 - Prob. 4SAQ Ch. 19 - Prob. 5SAQ Ch. 19 - Prob. 6SAQ Ch. 19 - Prob. 7SAQ Ch. 19 - Prob. 8SAQ Ch. 19 - Prob. 9SAQ Ch. 19 - Prob. 10SAQ

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