\(n_{hh}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(n_{Br_2}=\dfrac{16}{160}=0.1\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(0.1.........0.1\)
\(n_{CH_4}=0.2-0.1=0.1\left(mol\right)\)
\(\%V_{C_2H_4}=\%V_{CH_4}=\dfrac{0.1}{0.2}\cdot100\%=50\%\)
Em cung cấp thông tin đề câu b lại nhé !