Đáp án+Giải thích các bước giải:

`n_(H_2)(text{ tổng})=(3,7185)/(24,79)=0,15(mol)` 

`a)`

`PTHH(1):2C_2H_5OH+2Na->2C_2H_5ONa+H_2uparrow`

`PTHH(2):2CH_3COOH+2Na->2CH_3COONa+H_2uparrow`

`b)`

Đặt `n_(C_2H_5OH)=a(mol);n_(CH_3COOH)=b(mol)`

Có: `m_(hh)=15,2(g)`

`=>46a+60b=15,2(1)`

Theo `PTHH(1)`

`=>n_(H_2)(1)=n_(C_2H_5OH)*1/2=a*1/2=0,5a(mol)`

Theo `PTHH(2)`

`=>n_(H_2)(2)=n_(CH_3COOH)*1/2=b*1/2=0,5b(mol)`

Mà `n_(H_2)` tổng `=0,15(mol)`

`=>0,5a+0,5b=0,15(2)`

Từ `(1)` và `(2)`

$=>\begin{cases} 46a+60b=15,2\\0,5a+0,5b=0,15 \end{cases}$

$=>\begin{cases} n_{C_2H_5OH}=a=0,2(mol)\\n_{CH_3COOH}=b=0,1(mol) \end{cases}$

`=>m_(C_2H_5OH)=0,2*46=9,2(g)`

`=>m_(CH_3COOH)=15,2-9,2=6(g)`

`c)`

Theo `PTHH(1)`

`=>n_(C_2H_5ONa)=n_(C_2H_5OH)=0,2(mol)`

Theo `PTHH(2)`

`=>n_(CH_3COONa)=n_(CH_3COOH)=0,1(mol)`

`=>m_(text{muối})=0,2*68+0,1*82=21,8(g)`

$\color{#FFFF00}{Vi}\color{#CCFF00}{et}\color{#99FF00} {Na}\color{#66FF00}{m}\color{#33FF00}{20}\color{#00FF00} {10}$