Answer
$2.34\ g$
Work Step by Step
Number of moles of the acid:
$0.146\ M\times 250.\ mL=36.5\ mmol$
From stoichiometry:
$0.073\ mol$ of hydrazine
Mass required:
$0.073\ mol\times 32.04\ g/mol=2.34\ g$
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: 64
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