$a)\\ 2CH_3COOH+2Na\to 2CH_3COONa+H_2\\ 2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2\\ 2CH_3COOH+CaCO_3\to (CH_3COO)_2Ca+CO_2+H_2O\\ CH_3COOH+C_2H_5OH \buildrel{{H_2SO_4(dac),t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O\\ b)$

Trong mỗi phần: $\begin{cases} n_{CH_3COOH}:a(mol)\\n_{C_2H_5OH}:b(mol) \end{cases}$

Phần 1:

$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ \sum n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}+\dfrac{1}{2}n_{C_2H_5OH}\\\to \dfrac{1}{2}a+\dfrac{1}{2}b=0,25\ (1)$

Phần 2:

$n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ n_{CH_3COOH}=2n_{CO_2}=0,2(mol)\\  \to a=0,2(mol)\ (2)\\ (1)(2)\to b=0,3(mol)\\ m_{CH_3COOH(A)}=0,2.60.3=36(g)\\ m_{C_2H_5OH(A)}=0,3.46.3=41,4(g)\\ c)\\ CH_3COOH+C_2H_5OH\buildrel{{H_2SO_4(dac),t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}<n_{C_2H_5OH}\to tinh\ theo\ CH_3COOH\\ n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,2(mol)\\ m_{CH_3COOC_2H_5}=0,2.88.60\%=10,56(g)$