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Theo chiều từ trái sang, từ trên xuống nhé
\(C_2H_2+H_2\underrightarrow{t^o,Pd,PbCO_3}C_2H_4\)
\(C_2H_4+H_2O\underrightarrow{H^+,t^o}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
\(C_2H_2+C_2H_4\xrightarrow[t^o]{Pd\text{/}PdCO_3}C_2H_4\\ C_2H_4+H_2O\xrightarrow[H^+]{t^o}C_2H_5OH\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đặc\right)}}CH_3COOC_2H_5+H_2O\\ CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\\ 2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\\ CH_3COOC_2H_5+KOH\rightarrow CH_3COOK+C_2H_5OH\\ C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
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
(1) \(C_2H_2+H_2\underrightarrow{t^o,Pd}C_2H_4\)
(2) \(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)
(3) \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
(4) \(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
\(Fe_2O_3+6CH_3COOH\rightarrow2\left(CH_3COO\right)_3Fe+3H_2O\)
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
\(CH_3COOH+CH_3OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOCH_3+H_2O\)
\(6CH_3COOH+2Al\rightarrow2\left(CH_3COO\right)_3Al+3H_2\)
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, \(n_{C_2H_5OH}=n_{CO_2}=0,75\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,75.46=34,5\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{34,5}{0,8}=43,125\left(ml\right)\)
\(C_2H_4+H_2O\underrightarrow{H^+,t^o}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
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
a, \(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
b, \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
c, \(C_6H_{12}O_6+2AgNO_3+3NH_3\underrightarrow{t^o}C_5H_{11}O_5COONH_4+2Ag+2NH_4NO_3\)
d, \(\left(RCOO\right)_3C_3H_5+3NaOH\underrightarrow{t^o}3RCOONa+C_3H_5\left(OH\right)_3\)
e, \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
f, \(H_2SO_4+2CH_3COONa\rightarrow2CH_3COOH+Na_2SO_4\)
\(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
$C + O_2 \xrightarrow{t^o} CO_2$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$CO_2 + C \to 2CO$
$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + NaOH \to CH_3COONa + C_2H_5OH$
a, \(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\) ( đk : nhiệt độ )
b, \(C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\) ( Dk : Nhiệt độ kèm chất xúc tác là H2SO4 đặc )
c, \(2CH_3COOH+Na_2O\rightarrow2CH_3COONa+H_2O\)
d, \(C_6H_6+Br_2\rightarrow C_6H_5Br+HBr\) ( Chất xúc tác là bột Fe )
e, \(2CH_3COOH+Cu\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Cu+2H_2O\)
f, \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\uparrow\)
g, \(C_6H_{12}O_6+Ag_2O\rightarrow C_6H_{12}O_7+2Ag\) ( đk : khí NH3 )
h, \(C_6H_6+3Cl_2\rightarrow C_6H_6Cl_6\) ( đk : Ánh sáng )
j, \(2CH_3COO+H_2SO_4\rightarrow2CH_2COOH+SO_4\)
l, \(C_2H_6+Cl_2\rightarrow HCl+C_2H_5Cl\) ( DK : AS)
q, \(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)
Cái PTHH j em tìm hiểu lại nhé!