Đáp án:

$n_{H_2}=0,24$ mol 

$\Rightarrow n_{C_nH_{2n+1}CH_2OH}+n_{C_nH_{2n+1}COOH}+n_{H_2O}=2n_{H_2}=0,24.2=0,48$ 

Mà $n_{C_nH_{2n+1}COOH}=n_{H_2O}$ nên $n_{C_nH_{2n+1}CH_2OH} + n_{C_nH_{2n+1}COOH}+n_{C_nH_{2n+1}COOH}=0,48$ 

$\Rightarrow n_{C_nH_{2n+1}CH_2OH \text{bđ}} + n_{C_nH_{2n+1}COOH}=0,48$

$n_{C_nH_{2n+1}CH_2OH\text{bđ}}=\dfrac{9,6}{12n+2n+1+14+17}=\dfrac{9,6}{14n+32}$ mol 

Ta có $n_{C_nH_{2n+1}COOH } < n_{C_nH_{2n+1}CH_2OH\text{bđ}}$ 

$\Rightarrow 0,48 < 2n_{C_nH_{2n+1}CH_2OH\text{bđ}}$ 

$\Rightarrow \dfrac{9,6}{14n+32}>\dfrac{0,48}{2}$ 

$\Rightarrow n<0,6$ 

$\Rightarrow n=0$ 

Vậy ancol ban đầu là $CH_3OH$ 

$n_{CH_3OH\text{bđ}}=0,3$ mol 

Đặt $n_{CH_3OH}=x$ mol, $n_{HCOOH}=y$ mol 

$\Rightarrow \begin{cases} x+y=0,3\\ x+y+y=0,48\end{cases}$ 

$\Rightarrow \begin{cases} x=0,12\\ y=0,18\end{cases}$

$HCOOH+CH_3OH\buildrel{{H_2SO_4 đ, t^o}}\over\rightleftharpoons HCOOCH_3+H_2O$

Ta có $n_{CH_3OH}<n_{HCOOH}$

$\Rightarrow n_{HCOOCH_3\text{lí thuyết}}=0,12$ mol 

$\Rightarrow m=0,12.60.80\%=5,76$ gam