\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br_2}=\dfrac{32,4}{22,4}=\dfrac{81}{56}mol\Rightarrow n_{C_2H_2}=\dfrac{81}{112}mol\)
\(n_A=\dfrac{7,62}{22,4}=\dfrac{381}{1120}mol\)
\(\Rightarrow n_{CH_4}=\dfrac{381}{1120}-\dfrac{81}{112}=-\dfrac{429}{1120}\)
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