Đáp án:a, \({C_2}{H_4}\)

            b, \({C_2}{H_3}Cl\)  monome vinyl clorua

                \({C_2}{H_4}C{l_2}\) 1,2-dichloroethane

Giải thích các bước giải:

 a, \({n_{C{O_2}}} = 0,2\) và \({n_{{H_2}O}} = 0,3\)

\({n_{{H_2}O}} - {n_{C{O_2}}} = 0,1\). Suy ra A là ankan 

\(\bar C = \dfrac{{{n_{C{O_2}}}}}{{{n_{ankan}}}} = 2 \to {C_2}{H_4}\)

b, \(\begin{array}{l}
{C_2}{H_4} + C{l_2} \to {C_2}{H_3}Cl + HCl\\
{C_2}{H_4} + C{l_2} \to {C_2}{H_4}C{l_2}
\end{array}\)

\(\left\{ \begin{array}{l}
a + b = 0,026\\
62,5a + 99b = 2,475
\end{array} \right.\)

a=0,004 b=0,022

a, 

$n_{CO_2}=\dfrac{8,8}{44}=0,2 mol$

$n_{H_2O}=\dfrac{5,4}{18}=0,3 mol > n_{CO_2}$

$\to$ ankan $C_nH_{2n+2}$

$\Rightarrow n=\dfrac{n_{CO_2}}{n_A}=\dfrac{0,2}{0,3-0,2}=2$

Vậy A là $C_2H_6$

b,

$n_A=\dfrac{0,75}{30}=0,025 mol=n_B$

$\Rightarrow M_B=\dfrac{2,475}{0,025}=99$

$B: C_2H_{5-x}Cl_x$

$\Rightarrow 12.2+5-x+35,5x=99$

$\Leftrightarrow x=2$ ($C_2H_4Cl_2$)

CTCT: 

(1) $CH_3-CHCl_2$ (1,1-đicloetan)

(2) $CH_2Cl-CH_2Cl$ (1,2-đicloetan)