PTHH : $C_2H_4 +Br_2 \rightarrow C_2H_4Br_2$ Ta có: $nBr_2=0.0175 (mol)\rightarrow nC_2H_4=0.0175 (mol) \rightarrow mC_2H_4=0.49(g)$
Mà $n_{hỗn hợp}=\frac{3,36}{22,4}=0.15(mol) \rightarrow nC_2H_6=0.15-0.0175=0.1325(mol) \rightarrow mC_2H_6=3.975(g)$
$ \rightarrow m_{hỗn hợp} =4.465(g)$
% $m_{C_{2}H_{4}} $ $=10.97$% $\rightarrow$ %$m_{C_{2}H_{6}}= 89.03$%