Đáp án:
\(\begin{array}{l}
\% {V_{{C_2}{H_2}}} = 1,5\% \\
\% {V_{C{H_4}}} = 98,5\% \\
{m_{{C_2}{H_2}}} = 0,117g\\
{m_{C{H_4}}} = 4,8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{B{r_2}}} = 0,009mol\\
\to {n_{{C_2}{H_2}}} = \dfrac{1}{2}{n_{B{r_2}}} = 0,0045mol\\
\to {V_{{C_2}{H_2}}} = 0,1008l\\
\to \% {V_{{C_2}{H_2}}} = \dfrac{{0,1008}}{{6,72}} \times 100\% = 1,5\% \\
\to \% {V_{C{H_4}}} = 98,5\% \\
{m_{{C_2}{H_2}}} = 0,117g\\
{n_{C{H_4}}} = 0,3mol\\
\to {m_{C{H_4}}} = 4,8g
\end{array}\)