Khi thoát ra là khí $CH_4$ 

a,

$ n_{Br_2} = \dfrac{m}{M} = \dfrac{8}{160} = 0,05 (mol) \\ C_2H_4 + Br_2 \rightarrow C_2H_4Br_2 \\ n_{C_2H_4} = n_{Br_2} = 0,05 (mol) \\ V_{C_2H_4} = n . 22,4 = 0,05 . 22,4 = 1,12 (l) \\ \rightarrow V_A = V_{C_2H_4} + V_{CH_4} = 1,12 + 3,36 = 4,48 (l) $ 

b,

$ \%  V_{C_2H_4} = \dfrac{V_{C_2H_4}}{V_{hh}}.100\% = \dfrac{1,12}{4,48}.100 = 25 \% \\ \%V_{CH_4} = 100\% - \%  V_{C_2H_4} = 100 - 25 = 75 \% $ 

c,

$ n_{CH_4} = \dfrac{V}{22,4} = \dfrac{3,36}{22,4} = 0,15 (mol) $

$ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O \\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O \\ n_{O_2} = 2.n_{CH_4} + 3.n_{C_2H_4} = 2.0,15 + 3.0,05 = 0,45 (mol) \\ \rightarrow V_{O_2} = n . 22,4 = 0,45 . 22,4 = 10,08 (l) $

Đáp án:

\(\begin{array}{l}
a){V_{hh}} = 4,48l\\
b)\\
\% {V_{{C_2}{H_4}}} = 25\% \\
\% {V_{C{H_4}}} = 75\% \\
c){V_{{O_2}}} = 10,08l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
a)\\
{n_{B{r_2}}} = 0,05mol\\
{n_{C{H_4}}} = 0,15mol\\
 \to {n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,05mol\\
 \to {V_{{C_2}{H_4}}} = 1,12l\\
 \to {V_{hh}} = 3,36 + 1,12 = 4,48l
\end{array}\)

\(\begin{array}{l}
b)\\
\% {V_{{C_2}{H_4}}} = \dfrac{{1,12}}{{1,12 + 3,36}} \times 100\%  = 25\% \\
\% {V_{C{H_4}}} = 100\%  - 25\%  = 75\% 
\end{array}\)

\(\begin{array}{l}
c)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 2{n_{C{H_4}}} + 3{n_{{C_2}{H_4}}} = 0,45mol\\
 \to {V_{{O_2}}} = 10,08l
\end{array}\)