Câu 9:

Bảo toàn nguyên tố C:

$\rm n_C=n_{CO_2}=\dfrac{3,36}{22,4}=0,15\ (mol)$

Bảo toàn nguyên tố H:

$\rm n_H=2n_{H_2O}=\dfrac{3,6×2}{18}=0,4\ (mol)$

$\rm n_O=\dfrac{3-0,15×12-0,4}{16}=0,05\ (mol)$

Tỉ lệ: $\rm n_C:n_H:n_O=0,15:0,4:0,05=3:8:1$

Công thức nguyên: $\rm (C_3H_8O)_n$

$\rm n_{X}=n_{O_2}=\dfrac{1,6}{32}=0,05\ (mol)$

$\rm M_X=(12×3+8+16)×n=\dfrac{3}{0,05}=60\ (g/mol)$

$\rm\to n=1$

Vậy X có CTPT là: $\rm C_3H_8O$

$\rm b)$ 

Các CTCT của X là:

$\rm CH_3-CH_2-CH_2-OH\\CH_3-CH(OH)-CH_3\\C_2H_5-O-CH_3$

Câu 10:

Tất cả trừ Ag, Cu, $\rm Na_2SO_4$.

Phản ứng xảy ra:

$\rm CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2\\CH_3COOH+K\to CH_3COOK+\dfrac{1}{2}H_2\\2CH_3COOH+Mg\to (CH_3COO)_2Mg+H_2\\2CH_3COOH+Zn\to (CH_3COO)_2Zn+H_2\\CH_3COOH+NaOH\to CH_3COONa+H_2O\\CH_3COOH+KOH\to CH_3COOK+H_2O\\2CH_3COOH+MgO\to (CH_3COO)_2Mg+H_2O\\2CH_3COOH+ZnO\to (CH_3COO)_2Zn+H_2O\\2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O\\2CH_3COOH+K_2CO_3\to 2CH_3COOK+CO_2+H_2O\\CH_3COOH+C_2H_5OH\overset{H_2SO_4\ (dac),\ t^°}\leftrightarrows CH_3COOC_2H_5+H_2O$