Đáp án:

\(\begin{array}{l}
2)A\\
5)\,D\\
7)\,A\\
9)B\\
11)\,C\\
12)D\\
13)C
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2)\\
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_A} = 2{n_{{H_2}}} = 0,2\,mol\\
{M_A} = \dfrac{{6,4}}{{0,2}} = 32\,g/mol \Rightarrow 14n + 18 = 32 \Rightarrow n = 1\\
 \Rightarrow CTHH\,A:C{H_3}OH\\
5)\\
{n_{C{O_2}}} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2\,mol\\
{n_A} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,05\,mol\\
A:{C_n}{H_{2n + 1}}OH\\
n = \dfrac{{{n_{C{O_2}}}}}{{{n_A}}} = \dfrac{{0,15}}{{0,05}} = 3 \Rightarrow CTHH:{C_3}{H_7}OH\\
m = 0,05 \times 60 = 3g\\
7)\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2\,mol\\
{n_A} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,1\,mol\\
A:{C_n}{H_{2n + 1}}OH\\
n = \dfrac{{{n_{C{O_2}}}}}{{{n_A}}} = \dfrac{{0,1}}{{0,1}} = 1 \Rightarrow CTHH:C{H_3}OH(C{H_4}O)\\
9)\\
{C_n}{H_{2n + 1}}OH + \dfrac{{3n}}{2}{O_2} \to nC{O_2} + (n + 1){H_2}O\\
{n_X} = \dfrac{{5,328 \times 2}}{{3n}} = \dfrac{{3,552}}{n}\,mol\\
{M_X} = \dfrac{{81,696}}{{\dfrac{{3,552}}{n}}} = 23n\,g/mol\\
 \Rightarrow 14n + 18 = 23n \Rightarrow n = 2\\
CTHH:{C_2}{H_6}O\\
11)\\
{M_B} > {M_A} \Rightarrow B:ete\\
{M_B} = 2{M_A} - 18 \Leftrightarrow 2{M_A} - {M_B} = 18(1)\\
{d_{B/A}} = 1,4375 \Leftrightarrow \dfrac{{{M_B}}}{{{M_A}}} = 1,4375 \Rightarrow 1,4375{M_A} - {M_B} = 0(2)\\
(1),(2) \Rightarrow {M_A} = 32;{M_B} = 46\\
 \Rightarrow 14n + 18 = 32 \Rightarrow n = 1\\
 \Rightarrow A:C{H_3}OH\\
12)\\
{d_{X/Y}} = 1,32 \Leftrightarrow \dfrac{{14n + 18}}{{14n}} = 1,32\\
 \Rightarrow n = 4 \Rightarrow CTHH:{C_4}{H_{10}}O\\
13)\\
RC{H_2}OH + CuO \to RCHO + Cu + {H_2}O\\
{n_{RC{H_2}OH}} = {n_{RCHO}} \Leftrightarrow \dfrac{6}{{{M_R} + 31}} = \dfrac{{5,8}}{{{M_R} + 29}}\\
 \Rightarrow {M_R} = 29 \Rightarrow R:{C_2}{H_5}\\
CTPT:C{H_3}C{H_2}C{H_2}OH
\end{array}\)

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