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\(\begin{array}{l}
1)\\
\rm{(1)\quad CH_4 + Cl_2 \xrightarrow[khuếch\ tán]{ánh\ sáng} CH_3Cl + HCl\\
(2)\quad 2CH_4 \xrightarrow[làm\ lạnh\ nhanh]{1500^\circ C} C_2H_2 + 3H_2\\
(3)\quad C_2H_2 + H_2 \xrightarrow{t^\circ, \ Pd/PbCO_2} C_2H_4\\
(4)\quad nCH_2=CH_2 \xrightarrow{t^\circ,\ xt,\ p} (\kern-6pt-CH_2-CH_2-\kern-6pt)_n\\
(5)\quad CH\equiv CH + HCl \xrightarrow{t^\circ,\ HgCl_2} CH_2=CHCl\\
(6)\quad nCH_2=CHCl \xrightarrow{t^\circ,\ xt,\ p} (\kern-6pt-CH_2-CHCl-\kern-6pt)_n\\
(7)\quad C_2H_2 + 2Br_2 \longrightarrow C_2H_2Br_4}\\
2)\\
\text{Bảo toàn nguyên tố C:}\\
n_C = n_{CO_2} = \dfrac{13,2}{44} = 0,3\ \rm mol\\
m_C = 0,3\times 12 = 3,6\ g\\
\text{Bảo toàn nguyên tố H:}\\
n_H = 2n_{H_2O} = 2\times \dfrac{7,2}{18} = 0,8\ \rm mol\\
m_H = 0,8\times 1 = 0,8\ g\\
\text{Khối lượng O có trong A:}\\
m_O = m_A - (m_C + m_H) = 6 - (3,6 + 0,8) = 1,6\ g\\
n_O = \dfrac{1,6}{16} = 0,1\ \rm mol\\
\text{Lập tỉ lệ mol:}\\
n_C\ :\ n_H\ :\ n_O = 0,3\ :\ 0,8\ :\ 0,1 = 3\ :\ 8\ :\ 1\\
\Rightarrow \text{CTPT của A có dạng:}\ \rm (C_3H_8O)_n\\
\text{Ta lại có:}\\
\quad M_A = 60\\
\Leftrightarrow (12\times 3 + 8 + 16)\times n = 60\\
\Leftrightarrow 60n = 60\\
\Leftrightarrow n = 1\\
\Rightarrow A:\ \rm C_3H_8O\\
\text{CTCT có thể có của A:}\\
\begin{array}{|c|c|}
\hline
H\quad H\quad H\quad \quad\\
H-\mathop{C}\limits_{\Big|}^{\Big|}-\mathop{C}\limits_{\Big|}^{\Big|}-\mathop{C}\limits_{\Big|}^{\Big|}-O-H\\
H\quad H\quad H\quad \quad\\
\hline
H\quad H\quad H\\
H-\mathop{C}\limits_{\Big|}^{\Big|}-\mathop{C}\limits_{\Big|}^{\Big|}-\mathop{C}\limits_{\Big|}^{\Big|}-H\\
H\quad \mathop{O}\limits_{\Big|}\quad H\\
H\\
\hline
H\quad H\qquad \quad H\\
H-\mathop{C}\limits_{\Big|}^{\Big|}-\mathop{C}\limits_{\Big|}^{\Big|}-O-\mathop{C}\limits_{\Big|}^{\Big|}-H\\
H\quad H\qquad \quad H\\
\hline
\end{array}
\end{array}\)