Đáp án:
\(\begin{array}{l}
\% {V_{{C_2}{H_4}}} = 60\% \\
\% {V_{{C_2}{H_2}}} = 40\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
b)\\
{n_{hh}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{B{r_2}}} = \dfrac{{11,2}}{{160}} = 0,07mol\\
hh:{C_2}{H_4}(a\,mol),{C_2}{H_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,05\\
a + 2b = 0,07
\end{array} \right.\\
\Rightarrow a = 0,03mol;b = 0,02mol\\
\% {V_{{C_2}{H_4}}} = \dfrac{{0,03 \times 22,4}}{{1,12}} \times 100\% = 60\% \\
\% {V_{{C_2}{H_2}}} = 100 - 60 = 40\%
\end{array}\)