Em tham khảo nha :

\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
2AgN{O_3} + {C_2}{H_2} + 2N{H_3} \to 2N{H_4}N{O_3} + A{g_2}{C_2}\\
b)\\
{V_{{C_3}{H_8}}} = 0,84l\\
{n_{{C_3}{H_8}}} = \dfrac{{0,84}}{{22,4}} = 0,0375mol\\
{m_{{C_3}{H_8}}} = 0,0375 \times 44 = 1,65g\\
{n_{A{g_2}{C_2}}} = \dfrac{{12,12}}{{240}} = 0,0505mol\\
{n_{{C_2}{H_2}}} = {n_{A{g_2}{C_3}}} = 0,0505mol\\
{V_{{C_{2{H_2}}}}} = 0,0505 \times 22,4 = 1,1312l\\
{m_{{C_2}{H_2}}} = 0,0505 \times 26 = 1,313g\\
{n_{{C_2}{H_4}}} = \dfrac{{3,36 - 0,84 - 1,1312}}{{22,4}} = 0,062mol\\
{m_{{C_2}{H_4}}} = 0,062 \times 28 = 1,736g\\
\% {V_{{C_3}{H_8}}} = \dfrac{{0,84}}{{22,4}} \times 100\%  = 25\% \\
\% {V_{{C_2}H}}_2 = \dfrac{{1,1312}}{{3,36}} \times 100\%  = 33,7\% \\
\% {V_{{C_2}{H_4}}} = 100 - 25 - 33,7 = 41,3\% \\
\% {C_3}{H_8} = \dfrac{{1,65}}{{1,65 + 1,313 + 1,736}} \times 100\%  = 35,1\% \\
\% {C_2}{H_2} = \dfrac{{1,313}}{{1,65 + 1,313 + 1,736}} \times 100\%  = 27,9\% \\
\% {C_2}{H_4} = 100 - 35,1 - 27,9 = 37\% 
\end{array}\)