Answer
$0.500\ g$
Work Step by Step
Number of moles of $Br_2$:
$0.102\ M\times 27.85\ mL=2.84\ mmol$
From stoichiometry:
$2.84\ mmol$ of vitamin C
$0.00284\ mol\times 176.13\ g/mol=0.500\ g$
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 179f: 76
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