Đáp án:

Giải thích các bước giải:

- Phần 1:

$n_{H_2} = \dfrac{3,36}{22,4} = 0,15 (mol)$

PTHH: 

$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2(1)$

$2CH_3COOH + 2Na \rightarrow 2CH_3COONa + H_2(2)$

Theo PTHH $(1),(2): n_{CH_3COOH} + n_{C_2H_5OH} = 2n_{H_2} = 2.0,15 = 0,3 (mol)$

- Phần 2:

$n_{CO_2} = \dfrac{4,48}{22,4} = 0,2 (mol)$

PTHH: $NaHCO_3 + CH_3COOH \rightarrow CH_3COONa + CO_2 \uparrow + H_2O(3)$

Theo PTHH $(3): n_{CH_3COOH} = n_{CO_2} = 0,2 (mol)$

Mà $n_{CH_3COOH} + n_{C_2H_5OH} = 0,3 (mol)$

$\rightarrow n_{C_2H_5OH} = 0,3 - 0,2 = 0,1 (mol)$

Ban đầu có:

$n_{C_2H_5OH} = 0,1.2 = 0,2 (mol)$

$\rightarrow m_{C_2H_5OH} = 0,2.46 = 9,2 (g)$

$n_{CH_3COOH} = 0,2.2 = 0,4 (mol)$

$m_{CH_3COOH} = 0,4.60 = 24 (g)$

$\rightarrow m_{\text{hỗn hợp}} 24 + 9,2 = 33,2 (g)$

$\rightarrow \%m_{C_2H_5OH} = \dfrac{9,2}{33,2} .100\% = 27,7\%$

$\rightarrow \%m_{CH_3COOH} = 100\% - 27,7\% = 72,3\%$

\(\begin{array}{l}
n_{H_2}=\frac{3,36}{22,4}=0,15(mol)\\
n_{CO_2}=\frac{4,48}{22,4}=0,2(mol)\\
\rm Phan\,1:\\
2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2\\
2CH_3COOH+Na\to 2CH_3COONa+H_2\\
\text{Theo PT: }n_{C_2H_5OH}+n_{CH_3COOH}=2n_{H_2}=0,3(1)\\
\rm Phan\,2:\\
CH_3COOH+NaHCO_3\to CH_3COONa+CO_2+H_2O\\
\text{Theo PT: }n_{CH_3COOH}=n_{CO_2}=0,2(mol)\\
\to n_{C_2H_5OH}=0,3-0,2=0,1(mol)\\
\to \%m_{CH_3COOH}=\frac{0,2.60}{0,2.60+0,1.46}.100\%\approx 72,29\%\\
\to \%m_{C_2H_5OH}=100-72,29=27,71\%\\
\rm \to Mai\,thi\,toan\,huhu
\end{array}\)