Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Mg}=\frac{4,32}{24}=0,18(mol)$

$Mg+2CH_3COOH\to (CH_3COO)_2Mg+H_2$

Theo PT: $n_{CH_3COOH}=2n_{Mg}=0,36(mol)$

$\to C\%_{CH_3COOH}=\frac{0,36.60}{500}.100\%=4,32\%$

$b\big)$

Sửa: $120\to 12^o$

$C_2H_5OH+O_2\xrightarrow{men\,giấm}CH_3COOH+H_2O$

Theo PT: $n_{C_2H_5OH}=n_{CH_3COOH}=0,36(mol)$

$\to V_{C_2H_5OH}=\frac{0,36.46}{0,8}=20,7(ml)$

$\to V_{rượu}=\frac{20,7.100}{12}=172,5(ml)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
{n_{Mg}} = \dfrac{{4,32}}{{24}} = 0,18mol\\
{n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,36mol\\
{m_{C{H_3}COOH}} = 0,36 \times 60 = 21,6g\\
C{\% _{C{H_3}COOH}} = \dfrac{{21,6}}{{500}} \times 100\%  = 4,32\% \\
b)\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
{n_{{C_2}{H_5}OH}} = {n_{C{H_3}COOH}} = 0,36mol\\
{m_{{C_2}{H_5}OH}} = 0,36 \times 46 = 16,56g\\
{V_{{C_2}{H_5}OH}} = \dfrac{{16,56}}{{0,8}} = 20,7c{m^3} = 20,7ml\\
{V_r} = \dfrac{{{V_{{C_2}{H_5}OH}} \times 100}}{{{D_r}}} = \dfrac{{20,7 \times 100}}{{12}} = 172,5ml
\end{array}\)