Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{C{H_4}}} = 66,7\% \\
\% {V_{{C_2}{H_4}}} = 33,3\% \\
b)\\
{m_{{C_2}{H_4}}} = 5,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{hh}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
16a + 28b = 6
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
\% {V_{C{H_4}}} = \dfrac{{0,2 \times 22,4}}{{6,72}} \times 100\% = 66,7\% \\
\% {V_{{C_2}{H_4}}} = 100 - 66,7 = 33,3\% \\
b)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{V_{{C_2}{H_4}}} = 33,3\% \times 13,44 = 4,48l\\
{n_{{C_2}{H_4}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{m_{{C_2}{H_4}}} = 0,2 \times 28 = 5,6g
\end{array}\)