## Chemistry and Chemical Reactivity (9th Edition)

$107.5\ g$
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: $20.11\ mol\times 1_{CH_4}/3_{H_2}\times 16.04\ g/mol=107.5\ g$