Chemistry and Chemical Reactivity (9th Edition)

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 179c: 23


$107.5\ g$

Work Step by Step

Percent yield: $37\%=15\ g/T.Y. \times 100\%$ $T.Y.= 40.54\ g$ Number of moles of hydrogen: $40.54\ g/2.016\ g/mol=20.11\ mol$ From stoichiometry: $20.11\ mol\times 1_{CH_4}/3_{H_2}\times 16.04\ g/mol=107.5\ g$
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