a,
\(CH_4+Cl_2\underrightarrow{^{as}}HCl+CH_3Cl\)
Khí tham gia phản ứng là CH4
\(\Rightarrow\%_{CH4}=\frac{1,12.100}{2,24}=50\%\Rightarrow\%_{C2H4}=50\%\)
b,
\(n_{CH4}=n_{Cl2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(\Rightarrow m_{Cl2}=0,05.71=3,55\left(g\right)\)