\(n_A=\frac{1,4}{22,4}=0,0625\left(mol\right)\rightarrow\overline{M}=\frac{0,975}{0,0625}=14,8\)
Đặt a,b,c là số mol của CH4 ; C2H4 ; H2
\(n_A=\frac{3,36}{22,4}=0,15\left(mol\right)\rightarrow a+b+c=0,15\left(1\right)\)
\(\overline{M}=\frac{16a+28b+2c}{a+b+c}=14,8\)
\(\Leftrightarrow1,a+13,2b-12,8c=0\left(2\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(m_{tang}=m_{C2H4}\rightarrow28c=0,84\left(3\right)\)
Từ (1) , (2) và (3) \(\rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,02\\c=0,03\end{matrix}\right.\)
\(\rightarrow\%_{CH4}=66,67\%,\%_{C2H4}=13,33\%,\%_{H2}=20\%\)
