Em tham khảo nha:

\(\begin{array}{l}
6)\\
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{hh}} = \dfrac{{1,008}}{{22,4}} = 0,045\,mol\\
{n_{B{r_2}}} = 0,7 \times 0,1 = 0,07\,mol\\
hh:{C_2}{H_2}(a\,mol),{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,045\\
2a + b = 0,07
\end{array} \right.\\
 \Rightarrow a = 0,025;b = 0,02\\
{V_{{C_2}{H_2}}} = 0,025 \times 22,4 = 0,56l\\
{V_{{C_2}{H_4}}} = 0,02 \times 22,4 = 4,48l\\
b)\\
\% {V_{{C_2}{H_2}}} = \dfrac{{0,56}}{{1,008}} \times 100\%  = 55,56\% \\
\% {V_{{C_2}{H_4}}} = 100 - 55,56 = 44,44\% \\
c)\\
\% {m_{{C_2}{H_2}}} = \dfrac{{0,025 \times 26}}{{0,025 \times 26 + 0,02 \times 28}} \times 100\%  = 53,72\% \\
\% {m_{{C_2}{H_4}}} = 100 - 53,72 = 46,28\% \\
7)\\
{C_6}{H_6} + B{r_2} \to {C_6}{H_5}Br + HBr\\
{n_{{C_6}{H_5}Br}} = \dfrac{{47,1}}{{157}} = 0,3\,mol\\
{n_{{C_6}{H_6}}} = {n_{{C_6}{H_5}Br}} = 0,3\,mol\\
{m_{{C_6}{H_6}}} = 0,3 \times 78 = 23,4g
\end{array}\)

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