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\(C_2H_4+H_2\underrightarrow{t^o,Ni}C_2H_6\)
\(C_2H_6+Cl_2\underrightarrow{as}C_2H_5Cl+HCl\)
\(C_2H_5Cl+NaOH\rightarrow C_2H_5OH+NaCl\)
\(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(CH_3COONa+NaOH\underrightarrow{CaO,t^o}CH_4+Na_2CO_3\)
\(C_2H_4+H_2O\underrightarrow{H^+,t^o}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
C2H4 → C2H5OH → CH3COOH → CH3COOC2H5 → C2H5OH
(1) C2H4 + H2O \(\underrightarrow{axit}\) C2H5OH
(2) C2H5OH + O2 \(\xrightarrow[25^0-30^0C]{mengiam}\) CH3COOH + H2O
(3) CH3COOH + C2H5OH → CH3COOC2H5 + H2O
(4) CH3COOC2H5 + NaOH \(\underrightarrow{t^0}\) CH3COONa + C2H5OH
$C + O_2 \xrightarrow{t^o} CO_2$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$CO_2 + C \to 2CO$
$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + NaOH \to CH_3COONa + C_2H_5OH$
C2H4 + H2O \(\underrightarrow{lênmen}\) C2H5OH
C2H5OH + O2 \(\underrightarrow{lênmen}\) CH3COOH + H2O
CH3COOH + C2H5OH \(\rightarrow\) CH3COOC2H5 + H2O
CH3COOC2H5 + NaOH \(\rightarrow\) CH3COONa + C2H5OH
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, \(n_{C_2H_5OH}=n_{CO_2}=0,75\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,75.46=34,5\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{34,5}{0,8}=43,125\left(ml\right)\)
\(C_2H_2 + H_2 \xrightarrow{t^o,xt} C_2H_4\\ C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH\\ C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ CH_3COOH + NaOH \to CH_3COONa + H_2O\)
\(\left(-C_6H_{10}O_5-\right)_n+nH_2O\rightarrow nC_6H_{12}O_6\\ C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
\(C_2H_4+H_2O\underrightarrow{t^o,H^+}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\underrightarrow{t^o,H^+}CH_3COOC_2H_5+H_2O\)
\(CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\)
\(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)