\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{C_2H_4Br_2} = \dfrac{4,7}{199}= 0,025(mol)\\ n_{C_2H_6} = \dfrac{3-0,025.22,4}{22,4}= \dfrac{61}{560}(mol)\\ \Rightarrow m_{hh} = 0,025.28 + \dfrac{61}{560}30 \dfrac{1111}{280}\\ \Rightarrow \%m_{C_2H_4} = \dfrac{0,025.28}{ \dfrac{1111}{280}}.100\% = 17,64\%\\ \%m_{C_2H_6} = 100\% - 17,64\% = 82,86\%\)
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