a,
$n_{CO_2}=\dfrac{28,6}{44}=0,65(mol)$
X gồm 2 anken nên $n_X=n_{Br_2}=0,25(mol)$
$\Rightarrow \overline{C}=\dfrac{0,65}{0,25}=2,6$
Vậy 2 anken là $C_2H_4$, $C_3H_6$
b,
Gọi $x$, $y$ là số mol $C_2H_4$, $C_3H_6$
$\Rightarrow x+y=0,25$ $(1)$
Bảo toàn $C$: $2x+3y=0,65$ $(2)$
$(1)(2)\Rightarrow x=0,1; y=0,15$
$\to m=0,1.28+0,15.42=9,1g$
$\%m_{C_2H_4}=\dfrac{0,1.28.100}{9,1}=30,77\%$
$\to \%m_{C_3H_6}=69,23\%$
