Em tham khảo nha:
\(\begin{array}{l}
a)\\
{C_n}{H_{2n}} + B{r_2} \to {C_n}{H_{2n}}B{r_2}\\
{n_{B{r_2}}} = \dfrac{8}{{160}} = 0,05\,mol\\
{n_A} = {n_{B{r_2}}} = 0,05\,mol\\
{M_A} = \dfrac{{2,8}}{{0,05}} = 56g/mol \Rightarrow 14n = 56\\
\Rightarrow n = 4 \Rightarrow CTPT:{C_4}{H_8}\\
b)\\
C{H_3} - CH = CH - C{H_3}
\end{array}\)