Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677e: 93

Answer

a) $30\ mL$ b) $0.627\ g$

Work Step by Step

a) The number of moles of oxalic acid: $28\ g\times 1.2/100\div90.03\ g/mol=0.00373\ mol$ Oxalic acid is a diprotic acid, so two moles of NaOH is required to neutralize each mole of oxalic acid. $0.25\ M\times V = 2\times0.00373\ mol$ $V=0.030\ L = 30\ mL$ b) 1 mol of calcium oxalate for each mol of oxalic acid: $0.00373\ mol\times 168.18\ g/mol=0.627\ g$
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