17 D

18 A

19 C

20 D

 Vừa khóc vừa lm nên có j sai thì bảo tui nhan

Đáp án:

 17 C

18 B

19 B

20 A

Giải thích các bước giải:

\(\begin{array}{l}
17)\\
nC{O_2} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
n{O_2} = \dfrac{{5,6}}{{32}} = 0,175\,mol\\
CTPT\,X:{C_n}{H_{2n + 2}}{O_x}(1 \le x \le n)\\
{C_n}{H_{2n + 2}}{O_x} + \dfrac{{3n + 1 - x}}{2}{O_2} \to nC{O_2} + (n + 1){H_2}O\\
0,05 \times n = 0,15\\
 \Rightarrow n = 3\\
0,05 \times \dfrac{{3n + 1 - x}}{2} = 0,175\\
 \Rightarrow 3n + 1 - x = 7\\
 \Rightarrow x = 3\\
 \Rightarrow CT\,X:{C_3}{H_5}{(OH)_3}\\
18)\\
2{C_3}{H_7}OH + 2Na \to 2{C_3}{H_7}ONa + {H_2}\\
n{C_3}{H_7}OH = \dfrac{9}{{60}} = 0,15\,mol\\
n{H_2} = 0,075\,mol\\
V = 0,075 \times 22,4 = 1,68l\\
19)\\
n{H_2}O = \dfrac{{6,3}}{{18}} = 0,35\,mol\\
nC{O_2} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
nX = 0,35 - 0,25 = 0,1\,mol\\
\overline C  = \dfrac{{0,25}}{{0,1}} = 2,5\\
 \Rightarrow Y:{C_2}{H_5}OH\\
20)\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
n{C_6}{H_2}B{r_3}OH = \dfrac{{26,48}}{{331}} = 0,08\,mol\\
nB{r_2} = 0,08 \times 3 = 0,24\,mol\\
mB{r_2} = 0,24 \times 160 = 38,4g
\end{array}\)