+ Pứ với `Na_2CO_3`

`n_↓=10:100=0,1(mol)`

`PTHH:CO_2+Ca(OH)_2→CaCO_3+H_2O`

 `(mol)--0,1-----0,1--`

`PTHH:2CH_3COOH+Na_2CO_3→2CH_3COONa+H_2O+CO_2`

`(mol)--0,2--------0,1--`

Vậy `n_{CH_3COOH}=0,2(mol)`

+ Pứ với `Na`

`PTHH:2CH_3COOH+2Na→2CH_3COONa+H_2`

`(mol)--0,2---------0,1--`

Ta có: `n_{H_2}=4,48:22,4=0,2(mol)`

`⇒C_2H_5OH` pứ với Na sinh ra `0,2-0,1=0,1(mol)`

`PTHH:2C_2H_5OH+2Na→2C_2H_5ONa+H_2`

`(mol)--0,2---------0,1--`

`%m_{CH_3COOH}=\frac{0,2.60}{0,2.60+0,2.46}.100=56,6(%)`

`%m_{C_2H_5OH}=100-56,6=43,4(%)`

Đáp án:

 

Giải thích các bước giải:

-

$n_{CaCO3} = \frac{10}{100}  = 0,1(mol)$

$CO_2 + Ca(OH)_2 → CaCO_3 + H_2O$

$⇒ n_{CO2} = n_{CaCO3} = 0,1(mol)$

$2CH_3COOH + Na_2CO_3 → 2CH_3COONa + CO_2 + H_2O$

$⇒ n_{CH3COOH} = 2n_{CO2} = 0,1.2 = 0,2(mol)$

-

$n_{H2} = \frac{4,48}{22,4} = 0,2(mol)$

$2C_2H_5OH + 2Na → 2C_2H_5ONa + H_2$

$2CH_3COOH + 2Na → 2CH_3COONa + H_2$

$⇒ \frac{n_{CH3COOH} + n_{C2H5OH}}{2} = n_{H2} = 0,2$

$⇒ n_{C2H5OH} = 0,2(mol)$

Vậy :

$\%m_{C2H5OH} = \frac{0,2.46}{0,2.46+0,2.60} .100\% = 43,4\%$

$⇒ \$m_{CH3COOH} = 100\% - 43,4\% = 56,6\%$