Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
{n_{C{O_2}}} = \dfrac{{16,8}}{{22,4}} = 0,75\,mol\\
{n_{{H_2}O}} = \dfrac{{16,2}}{{18}} = 0,9\,mol\\
{n_{C{O_2}}} < {n_{{H_2}O}} \Rightarrow X:ankan\\
{n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,9 - 0,75 = 0,15\,mol\\
C = \dfrac{{{n_{C{O_2}}}}}{{{n_X}}} = \dfrac{{0,75}}{{0,15}} = 5 \Rightarrow CTPT:{C_5}{H_{12}}\\
b)\\
{C_5}{H_{12}} + C{l_2} \to {C_5}{H_{11}}Cl + HCl\\
{C_5}{H_{12}} + 8{O_2} \xrightarrow{t^0} 5C{O_2} + 6{H_2}O\\
{C_5}{H_{12}} \xrightarrow{t^0,xt} {C_4}{H_8} + C{H_4}\\
2)\\
C{H_2} = CH - C{H_2} - C{H_2} - C{H_2} - C{H_3}:h{\rm{ex}} - 1 - en\\
C{H_3} - CH = CH - C{H_2} - C{H_2} - C{H_3}:h{\rm{ex}} - 2 - en\\
C{H_3} - C{H_2} - CH = CH - C{H_2} - C{H_3}:h{\rm{ex}} - 3 - en\\
C{H_2} = C(C{H_3}) - C{H_2} - C{H_2} - C{H_3}:2 - metyl\,pent - 1 - en\\
C{H_2} = CH - CH(C{H_3}) - C{H_2} - C{H_3}:3 - metyl\,pent - 1 - en
\end{array}\)

 

Câu 1:

a,

$n_{CO_2}=\dfrac{16,8}{22,4}=0,75(mol)$

$n_{H_2O}=\dfrac{16,2}{18}=0,9(mol)$

$n_{CO_2}<n_{H_2O}\to X$ là ankan $C_nH_{2n+2}$

$n_X=n_{H_2O}-n_{CO_2}=0,15(mol)$

$\to n=\dfrac{n_{CO_2}}{n_X}=5$

Vậy CTPT $X$ là $C_5H_{12}$

b,

(1) $C_5H_{12}+Cl_2\xrightarrow{{as}} C_5H_{11}Cl+HCl$

(2) $C_5H_{12}+8O_2\xrightarrow{{t^o}} 5CO_2+6H_2O$

(3) không phản ứng 

Câu 2:

$CH_2=CH-CH_2-CH_2-CH_2-CH_3$: hex-1-en 

$CH_3-CH=CH-CH_2-CH_2-CH_3$: hex-2-en 

$CH_3-CH_2-CH=CH-CH_2-CH_3$: hex-3-en 

$CH_2=C(CH_3)-CH_2-CH_3$: 2-metylbut-1-en 

$CH_3-CH(CH_3)-CH=CH_2$: 3-metylbut-1-en