Đáp án:

\(\begin{array}{l}
b,\\
\% {V_{{C_2}{H_2}}} = 25\% \\
\% {V_{C{H_4}}} = 75\% \\
c,{V_{{O_2}}} = 9,52l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a,{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
b,\\
{n_{B{r_2}}} = 0,1mol\\
 \to {n_{{C_2}{H_2}}} = \dfrac{1}{2}{n_{B{r_2}}} = 0,05mol\\
 \to {V_{{C_2}{H_2}}} = 1,12l\\
 \to {V_{C{H_4}}} = 3,36l\\
 \to \% {V_{{C_2}{H_2}}} = \dfrac{{1,12}}{{4,48}} \times 100\%  = 25\% \\
 \to \% {V_{C{H_4}}} = 100\%  - 25\%  = 75\% 
\end{array}\)

\(\begin{array}{l}
c,\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
2{C_2}{H_2} + 5{O_2} \to 4C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 2{n_{C{H_4}}} + \dfrac{5}{2}{C_2}{H_2} = 0,425mol\\
 \to {V_{{O_2}}} = 9,52l
\end{array}\)

a) PTHH:

  $C_2H_2+2Br_2→C_2H_2Br_4$

  $CH_4$ không phản ứng với dung dịch Brom.

b) $n_{Br_2}=\dfrac{16}{160}=0,1(mol)$

 $n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=\dfrac{1}{2}×0,1=0,05(mol)$

Mà ta lại có: $n_{\text{hh khí}}=\dfrac{4,48}{22,4}=0,2(mol)$ → $n_{CH_4}=0,2-0,05=0,15(mol)$

⇒ %$V_{C_2H_2}=\dfrac{0,05}{0,2}×100$% $=$ $25$%

⇒ %$V_{CH_4}=100$% $-$ $25$% $=$ $75$%

c) $CH_4+2O_2\buildrel{{t^o}}\over\longrightarrow CO_2+2H_2O$

    $2C_2H_2+5O_2\buildrel{{t^o}}\over\longrightarrow 4CO_2+2H_2O$

Ta có: 

 $n_{O_2}=2n_{CH_4}=2×0,15=0,3(mol)$

 $n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=\dfrac{5}{2}×0,05=0,125(mol)$

⇒ $V_{O_2}=22,4×(0,3+0,125)=9,52(l)$