Answer
C
Work Step by Step
Theoretical yield:
$125\ g\ Al_4C_3\div 143.96\ g\ Al_4C_3/mol\ Al_4C_3\times \dfrac{3\ mol\ CH_4}{1\ mol\ Al_4C_3}\times16.04\ g\ CH_4/mol\ CH_4=41.78\ g\ CH_4$
Percent yield:
$13.6\div41.78\times100\%=32.55\%$
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 - 4-3 Percent Yield - Review & Check for Section 4-3 - Page 149: 2
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