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\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2o\\ CH_3COOH+C_2H_5OH⇌\left(xt,t^o\right)CH_3COOC_2H_5+H_2O\\ 2CH_3COOC_2H_5+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2C_2H_5OH\)
\(a) C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH\\ C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ CH_3COOH + C_2H_5OH \to CH_3COOC_2H_5 + H_2O\\ b) C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+}C_6H_{12}O_6 + C_6H_{12}O_6\\ C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH\\C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ CH_3COOH + NaOH \to CH_3COONa + H_2O\\ c) CH_3COOC_2H_5 + NaOH \to CH_3COONa + C_2H_5OH\\ d) 2CH_3COOC_2H_5 + Ca(OH)_2 \to (CH_3COO)_2Ca + 2C_2H_5OH\)
\((CH_3COO)_2Ca + H_2SO_4 \to CaSO_4 + 2CH_3COOH\)
(1) 2C2H5OH + O2 \(\underrightarrow{mengiam}\) 2CH3COOH + 2H2O
(2) CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O (xt: axit, to )
(3) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
CH3COONa + HCl → CH3COOH + NaCl
\(C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH\\ C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH\\ 2CH_3COOH + Ca(OH)_2 \to (CH_3COO)_2Ca + 2H_2O\)
C6H12O6 => C2H5OH + 2CO2
C2H5OH+O2 
CH3-COOH+H2O
CH3COOH+C2H5OH->CH3COOC2H5+H2O(môi trường H2SO4 đặc , nhiệt độ)
CH3COOC2H5+H2O->CH3COOH+C2H5OH(môi trường H2SO4 đặc , nhiệt độ)
2CH3COOH + Ca(OH)2 → (CH3COO)2Ca + 2H2O
1)
Hai phương pháp là:
1 Dùng quỳ tím: Axit axetic làm quỳ tím hóa đỏ Rượu etylic không làm quỳ tím đổi màu
2 Dùng Na2CO3 hoặc CaCO3: CH3COOH cho khí CO2 thoát ra C2H5OH không có phản ứng
2)
\(C_2H_4+H_2O\rightarrow C_2H_5OH\)
\(C_2H_5OH+O_2\rightarrow CH_3COOH+H_2O\)
\(C_2H_5OH+CH_3COOH⇌H_2O+CH_3COOC_2H_5\)
(1) \(C_6H_{12}O_6\underrightarrow{t^o,xt}2C_2H_5OH+2CO_2\)
(2) \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
(3) \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
(4) \(CH_3COONa+NaOH\underrightarrow{^{CaO,t^o}}CH_4+Na_2CO_3\)
\(C_2H_4+H_2O\underrightarrow{^{170^0C,H_2SO_4}}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{^{\text{men giấm}}}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(Đk:H_2SO_{4\left(đ\right)},t^0\right)\)
\(2CH_3COOC_2H_5+Ca\left(OH\right)_2\underrightarrow{^{t^0}}\left(CH_3COO\right)_2Ca+2C_2H_5OH\)
\(\left(CH_3COO\right)_2Ca+Na_2CO_3\rightarrow2CH_3COONa+CaCO_3\downarrow\)
\(C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH\\ C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ CH_3COOH + C_2H_5 \rightleftharpoons CH_3COOC_2H_5 + H_2O\\ 2CH_3COOC_2H_5 + Ca(OH)_2 \to (CH_3COO)_2Ca + 2C_2H_5OH\\ (CH_3COO)_2Ca + Na_2CO_3 \to CaCO_3 + 2CH_3COONa\)