Đáp án:
\(\left\{ \begin{gathered}
{C_2}{H_4} \hfill \\
{C_3}{H_6} \hfill \\
\end{gathered} \right.\)
Giải thích các bước giải:
$n_{Br_2}=\dfrac{64}{160}=0,4\ mol$
$\text{PTHH:}$
\({C_{\overline n }}{H_{2\overline n }} + {\text{ }}B{r_2} \to {C_{\overline n }}{H_{2\overline n }}B{r_2}\)
$\text{Theo PTHH:}\ {n_{{C_{\overline n }}{H_{2\overline n }}}} = {n_{B{r_2}}} = 0,4\ mol$
\( \to {M_{{C_{\overline n }}{H_{2\overline n }}}} = \dfrac{{14}}{{0,4}} \to \overline n = 2,5 \to \left\{ \begin{gathered}
{C_2}{H_4} \hfill \\
{C_3}{H_6} \hfill \\
\end{gathered} \right.\)