Answer
$0.33\ g$
Work Step by Step
Number of moles of the acid:
$50.0\ mL \div 0.125\ M=6.25\ mmol$
From stoichiometry, number of moles of acid required:
$0.00313\ mol$
Mass required:
$0.00313\ mol\times 105.99\ g/mol=0.33\ g$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179e: 62
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