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C2H5OH + O2 ---men giấm--> CH3COOH + H2O
2CH3COOH + 2Na ----> 2CH3COONa + H2
b.
CaC2 + 2H2O ---> C2H2 + Ca(OH)2
C2H2 + H2 -xt,to--> C2H4
C2H4 + H2O ---> C2H5OH
2C2H5OH + Na ---> 2C2H5ONa + H2
2CH3COOH+Na->CH3COONa+H2
CH3COOH+Na2CO3->CH2COONa+h2O+CO2
2CH3COOH+CaCO3->(CH3COO)2Ca+H2O+CO2
2CH3COOH+Mg->(CH3COO)2Mg+H2
Bài 4: Hoàn thành các PTHH sau:
a, Na + CH3COOH → CH3COONa + \(\dfrac{1}{2}\)H2
b, Na2CO3 + 2CH3COOH → 2CH3COONa + H2O + CO2
c, 2CH3COOH + CaCO3 → (CH3COO)2Ca + H2O + CO2↑
d, 2CH3COOH + Mg → (CH3COO)2Mg + H2↑
a) \(2Na+2CH_3COOH\rightarrow2CH_3COONa+H_2\)
b) \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2+H_2O\)
c) \(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
d) \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
\(a) 2Na+2CH_3COOH→2CH_3COONa+H_2 \)
\(b) Na_2CO_3+2CH_3COOH→2CH_3COONa+CO_2+H_2O \)
\(c) 2CH_3COOH+CaCO_3→(CH_3COO)_2Ca+H_2O+CO_2↑ \)
\(d) 2CH_3COOH+Mg→(CH_3COO)_2Mg+H_2↑\)
1)
a)
$C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
b)
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
2)
a) $n_{CO_2} = \dfrac{16,8}{22,4} = 0,75(mol)$
$C_6H_{12}O_6 \xrightarrow{men\ rượu} 2CO_2 + 2C_2H_5OH$
$n_{glucozo} = \dfrac{1}{2}n_{CO_2} = 0,375(mol)$
$m_{glucozo} = 0,375.180 = 67,5(gam)$
b) $n_{C_2H_5OH} = n_{CO_2} = 0,75(mol)$
$m_{C_2H_5OH} = 0,75.46 = 34,5(gam)$
$V_{C_2H_5OH} = \dfrac{34,5}{0,8}= 43,125(ml)$
Câu 1:
a, \(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O\)
\(CH_3COOH+Na\rightarrow CH_3COOH+\dfrac{1}{2}H_2\)
b, \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)
Câu 2:
a, \(n_{CO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
\(C_6H_{12}O_6\underrightarrow{t^o,xt}2C_2H_5OH+2CO_2\)
Theo PT: \(n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,375\left(mol\right)\)
\(\Rightarrow m_{C_6H_{12}O_6}=0,375.180=67,5\left(g\right)\)
b, \(n_{C_2H_5OH}=n_{CO_2}=0,75\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,75.46=34,5\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{34,5}{0,8}=43,125\left(ml\right)\)
a)
C2H4 + H2O \(\xrightarrow{t^o,xt}\) C2H5OH
C2H5OH + O2 \(\xrightarrow{men\ giấm}\) CH3COOH + H2O
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
CH3COOC2H5 + KOH → CH3COOK + C2H5OH
b)
(1) 2C2H5OH + 2Na → 2C2H5ONa + H2
(2) 2CH3COOH + Mg → (CH3COO)2Mg + H2
(3) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
(4) (RCOO)3C3H5 + 3H2O ⇌ 3RCOOH + C3H5(OH)3
C2H4 + H2O - 1700C , H2SO4-> C2H5OH
C2H5OH + O2 -men giấm-> CH3COOH + H2O
CH3COOH + C2H5OH <-H2SO4đ, to-> CH3COOC2H5 + H2O
CH3COOC2H5 + NaOH => CH3COONa + C2H5OH
\(CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\\ C_2H_2+H_2\xrightarrow[t^o]{Pd}C_2H_4\\ C_2H_4+H_2O\xrightarrow[t^o]{axit}C_2H_5OH\\ C_2H_5OH+O_2\underrightarrow{men,giấm}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
\(CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\)
\(C_2H_2+H_2\rightarrow\left(t^o,Pd\right)C_2H_4\)
\(C_2H_4+H_2O\rightarrow\left(t^o,axit\right)C_2H_5OH\)
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\rightarrow\left(t^o,H_2SO_4\left(đ\right)\right)CH_3COOC_2H_5+H_2O\)
Theo chiều từ trái sang, từ trên xuống nhé
\(C_2H_2+H_2\underrightarrow{t^o,Pd/PbCO_3}C_2H_4\)
\(C_2H_4+H_2O\underrightarrow{H^+,t^o}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(2CH_3COOH+MgO\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)
\(C_2H_2+H_2\xrightarrow[t^o]{Pd}C_2H_4\\ C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\\ C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_4đặc}CH_3COOC_2H_5+H_2O\\ CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ 2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
CH3COOH + NaOH → CH3COONa + H2O
https://phuongtrinhhoahoc.com/phuong-trinh/CH3COOH+Na2CO3/CH3COONa+H2O+CO2https://phuongtrinhhoahoc.com/phuong-trinh/CH3COOH+Na2CO3/CH3COONa+H2O+CO2