$a)$

$n_{CO_2}=\dfrac{3.9664}{24.79}=0.16(mol)$

$n_{H_2O}=\dfrac{3.96}{18}=0.22(mol)$

`->`$n_{H_2O}>n_{CO_2}$

`=>`$Alcohol$ no, đơn chức, mạch hở.

`=>`$n_{\text{hỗn hợp alcohol}}=n_{H_2O}-n_{CO_2}=0.06(mol)$

`=>`$C_{tb}=\dfrac{0.16}{0.06}=\dfrac{8}{3}≈2.67$

Mà hai $alchohol$ liên tiếp trong dãy đồng đẳng.

`=>`$C_2H_6O(x \,\, mol)$ và $C_3H_8O(y \,\, mol)$

$\begin{cases} PT\,\, tổng\,\, số\,\, mol\,\, \\PT\,\, bảo\,\, toàn\,\, nguyên\,\, tố\,\, carbon \end{cases}$

`->`$\begin{cases} x+y=0.06 \\2x+3y=0.16 \end{cases}$

`=>`$\begin{cases} x=0.02\\y=0.04  \end{cases}$

`=>`$a=46x+60y=3.32(gam)$

$b)$

`@`$C_2H_6O$

$\rm H_3C-CH_2-OH: \text{Ethanol} $

`@`$C_3H_8O$

$\rm H_3C-CH_2-CH_2-OH: \text{1-propanol}$

$\rm H_3C-CH(CH_3)-OH:\text{2-propanol }$

$c)$

$2C_nH_{2n+2}O \xrightarrow{t^o,H^+}C_nH_{2n+1}OC_nH_{2n+1}+H_2O$

Bảo toàn khối lượng:

$m=m_{ether}=m_{\text{hỗn hợp alcohol}}-m_{H_2O}$

$=a-18 \times \dfrac{1}{2} \times n_{\text{hỗn hợp alcohol}}$

$=3.32-18 \times \dfrac{0.06}{2}$

$=2.78(gam)$

Biết $\%H=80\%$`=>`$m_{thực}=2.78 \times 80\%=2.224(gam)$

`a)`

Alcohol chung: `C_nH_{2n+1}OH`

`n_{CO_2}={3,9664}/{24,79}=0,16(mol)`

`n_{H_2O}={3,96}/{18}=0,22(mol)`

`->n_{C_nH_{2n+1}OH}=n_{H_2O}-n_{CO_2}=0,06(mol)`

`->n={0,16}/{0,06}=2,667`

`->` 2 alcohol là `C_2H_5OH,C_3H_7OH`

`Hh:C_2H_5OH:x(mol);C_3H_7OH:y(mol)`

`->x+y=0,06;2x+3y=0,16`

`->x=0,02;y=0,04`

`a=0,02.46+0,04.60=3,32(g)`

`b)`

`C_2H_5OH:CH_3-CH_2-OH`

`C_3H_7OH:`

`CH_3-CH_2-CH_2-OH`

`CH_3-CH(OH)-CH_3`

`c)`

`2C_2H_5OH` $\xrightarrow{140^oC,H_2SO_4}$ `C_2H_5OC_2H_5+H_2O`

`2C_3H_7OH` $\xrightarrow{140^oC,H_2SO_4}$ `C_3H_7OC_3H_7+H_2O`

`n_{C_2H_5OC_2H_5\ l t}=1/2n_{C_2H_5OH}=0,01(mol)`

`n_{C_3H_7OC_3H_7\ l t}=1/2n_{C_3H_7OH}=0,02(mol)`

`m=(0,01.74+0,02.102).80\%=2,224(g)`