Đáp án:

\(\begin{array}{l}
a)\\
\% {V_{{C_2}{H_2}}} = 66,7\% \\
\% {V_{C{H_4}}} = 33,3\% \\
b)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
c)\\
{m_{B{r_2}}} = 32g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{m_{{C_2}{H_2}}} = 2,6g\\
{n_{{C_2}{H_2}}} = \dfrac{{2,6}}{{26}} = 0,1mol\\
{V_{{C_2}{H_2}}} = 0,1 \times 22,4 = 2,24l\\
\% {V_{{C_2}{H_2}}} = \dfrac{{2,24}}{{3,36}} \times 100\%  = 66,7\% \\
\% {V_{C{H_4}}} = 100 - 66,7 = 33,3\% \\
b)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
c)\\
{n_{B{r_2}}} = 2{n_{{C_2}{H_2}}} = 0,2mol\\
{m_{B{r_2}}} = 0,2 \times 160 = 32g
\end{array}\)

`n_{\text{hh khí}}=\frac{3,36}{22,4}=0,15(mol)`

`a)` `m_{\text{tăng}}=m_{C_2H_2}=2,6g`

`=> n_{C_2H_2}=\frac{2,6}{26}=0,1(mol)`

`=> n_{CH_4}=0,15-0,1=0,05(mol)`

`%V_{C_2H_2}=\frac{0,1.100%}{0,15}\approx 66,7%`

`%V_{CH_4}=\frac{0,05.100%}{0,15}\approx 33,3%`

`b)` Phương trình:

`CH≡CH+2Br_2\to CHBr_2-CHBr_2`

`c)` Ta nhận thấy: `n_{Br_2}=2n_{C_2H_2}=0,2(mol)`

`=> m_{Br_2}=0,2.160=32g`