General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 4 - Chemical Reactions - Questions and Problems - Page 173: 4.91

Answer

$325.3mL$

Work Step by Step

We know that $82.0g\space NaHCO_3\times \frac{1mol\space NaHCO_3}{84.01g\space NaHCO_3}\times \frac{1mol\space H_2SO_4}{2mol\space NaHCO_3}\times \frac{1L\space sol}{0.150mol\space H_2SO_4}=0.3253L=325.3mL$
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