Answer
C
Work Step by Step
Theoretical yield:
$125\ g\ Al_4C_3\div 143.96\ g\ Al_4C_3/mol\ Al_4C_3\times \dfrac{3\ mol\ CH_4}{1\ mol\ Al_4C_3}\times16.04\ g\ CH_4/mol\ CH_4=41.78\ g\ CH_4$
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