\(\begin{array}{l}
1)\\
C_2H_5OH+3O_2\xrightarrow{t^o}2CO_2+3H_2O\\
C_nH_{2n}O_2+\frac{3n-1}{2}O_2\xrightarrow{t^o}nCO_2+nH_2O\\
CO_2+2NaOH\to Na_2CO_3+H_2O(1)\\
Na_2CO_3+CO_2+H_2O\to 2NaHCO_3(2)\\
2NaHCO_3+CaCl_2\to CaCO_3+2NaCl+CO_2+H_2O(3)\\
NaHCO_3+Ca(OH)_2\to CaCO_3+NaOH+H_2O(4)\\
2)\\
m_{NaOH\,tang}=m_{CO_2}+m_{H_2O}=4,08(g)\\
\sum m_{CaCO_3}=m_1+m_2=6(g)\\
\to n_{CaCO_3}=\frac{6}{100}=0,06(mol)\\
Theo\,PT:\,\sum n_{CO_2}=0,5n_{NaHCO_3(3)}+n_{NaHCO_3(4)}=n_{CaCO_3}=0,06(mol)\\
\to n_{H_2O}=\frac{4,08-0,06.44}{18}=0,08(mol)\\
n_{ancol}=n_{C_2H_5OH}=n_{H_2O}-n_{CO_2}=0,02(mol)\\
\to m_A=m-0,02.46=m-0,92(g)\\
\to n_A.(14n+32)=m-0,92(g)\\
Theo\,PT:\,n_A.n+2n_{C_2H_5OH}=n_{CO_2}\\
\to n_A.n=0,02(mol)\\
Ta\,co\,m=m_A+m_{C_2H_5OH}=1,2+32n_A\\
Mat\,khac:\,37\%<\%m_A=\frac{m-0,92}{m}.100\%<42\%\\
\to 0,37<\frac{m-0,92}{m}<0,42\\
\to 0,37<\frac{n_A(14n+32)}{m}<0,42\\
\to 0,37<\frac{0,28+32n_A}{m}<0,42\\
\to 0,37<\frac{0,28+32n_A}{1,2+32n_A}<0,42\\
Xet\,0,37<\frac{0,28+32n_A}{1,2+32n_A}\\
\to 0,444+11,84n_A<0,28+32n_A\\
\to n_A>0,0081\\
\to n<\frac{0,02}{0,0081}=2,47\\
Xet\,\frac{0,28+32n_A}{1,2+32n_A}<0,42\\
\to 0,28+32n_A<0,504+13,44n_A\\
\to n_A<0,012\\
\to n>\frac{0,02}{0,012}=1,67\\
\to 1,67<n<2,47\\
\to n=2\,(n\in\mathbb{N^*})\\
\to A:C_2H_4O_2/CH_3COOH/axit\,axetic\\
3)\\
Bao\,toan\,C:\,
n_{CH_3COOH}=\frac{n_{CO_2}-2n_{C_2H_5OH}}{2}=0,01(mol)\\
CH_3COOH+C_2H_5OH\buildrel{{t^o,H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O\\
Vì\,n_{CH_3COOH}<n_{C_2H_5OH}\to H\,tinh\,theo\,CH_3COOH\\
n_{CH_3COOC_2H_5(lt)}=n_{CH_3COOH}=0,01(mol)\\
\to H=\frac{0,528}{0,01.88}.100\%=60\%
\end{array}\)