Đáp án:

\(\begin{array}{l}
b)\\
\% {V_{{C_2}{H_2}}} = 25\% \\
\% {C_3}{H_6} = 50\% \\
\% C{H_4} = 25\% \\
c)\\
{m_{B{r_2}}} = 19,2g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to A{g_2}{C_2} + 2N{H_4}N{O_3}\\
{C_3}{H_6} + B{r_2} \to {C_3}{H_6}B{r_2}\\
b)\\
{n_{A{g_2}{C_2}}} = \dfrac{{7,2}}{{240}} = 0,03mol\\
{n_{{C_2}{H_2}}} = {n_{A{g_2}{C_2}}} = 0,03mol\\
{n_{{C_3}{H_6}}} = \dfrac{{2,52}}{{42}} = 0,06mol\\
{n_{hh}} = \dfrac{{2,688}}{{22,4}} = 0,12mol\\
\% {V_{{C_2}{H_2}}} = \dfrac{{0,03}}{{0,12}} \times 100\%  = 25\% \\
\% {C_3}{H_6} = \dfrac{{0,06}}{{0,12}} \times 100\%  = 50\% \\
\% C{H_4} = 100 - 25 - 50 = 25\% \\
c)\\
{C_3}{H_6} + B{r_2} \to {C_3}{H_6}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{B{r_2}}} = {n_{{C_3}{H_6}}} + 2{n_{{C_2}{H_2}}} = 0,12mol\\
{m_{B{r_2}}} = 0,12 \times 160 = 19,2g
\end{array}\)