Hoan thành chuoi phải ưng: a) C2H4 --1--> C6H2OH--2--> CH3COOH --3--> ( CH3COO)2Fe / b) (-C6H10O5-)n --1--> C6H12O6 --2--> C2H5OH --3--> CH3COOH --4--> CH3COOC2H5
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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
23 tháng 4 2023
Chuỗi 1:
\(\left(1\right)CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ \left(2\right)CaO+3C\rightarrow\left(2000^oC,lò.điện\right)CaC_2+CO\uparrow\\ \left(3\right)CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\\ \left(4\right)C_2H_2+H_2\rightarrow\left(Ni,t^o\right)C_2H_4\\ \left(5\right)C_2H_4+H_2O\rightarrow\left(t^o,H^+\right)C_2H_5OH\\ \left(6\right)C_2H_5OH+2NaOH+CH_3COOH\rightarrow CH_3COONa+C_2H_5ONa+2H_2O\)
MN
27 tháng 4 2019
a) CaC2 + 2H2O --> Ca(OH)2 + C2H2
C2H2 + H2 -Ni-> C2H4
C2H4 + H2O --> C2H5OH
C2H5OH + O2 -mg-> CH3COOH + H2O
CH3COOH + C2H5OH <-H2SO4đ,to-> CH3COOC2H5 + H2O
b) (-C6H10O5-)n + nH2O -axit-> nC6H12O6
C6H12O6 -mr-> 2C2H5OH + 2CO2
C2H5OH + O2 -mg-> CH3COOH + H2O
CH3COOH + C2H5OH <-H2SO4đ,to-> CH3COOC2H5 + H2O
CH3COOC2H5 + NaOH -to-> CH3COONa + C2H5OH
27 tháng 4 2019
a/ CaC2 + 2H2O => Ca(OH)2 + C2H2
C2H2 + H2 => (Pd,to) C2H4
C2H4 + H2O => (140oC,H2SO4đ) C2H5OH
C2H5OH + O2 => (men giấm) CH3COOH + H2O
CH3COOH + C2H5OH => (H2SO4đ,to,pứ hai chiều) CH3COOC2H5 + H2O: pứ este hóa
19 tháng 4 2022
\(a,-\left(-C_6H_{10}O_5-\right)-_n+nH_2O\rightarrow nC_6H_{12}O_6\\ C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\uparrow\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_{4\left(đ\right)}]{t^o}CH_3COOC_2H_5+H_2O\)
\(b,C_{12}H_{22}O_{11}+H_2O\xrightarrow[t^o]{H^+}C_6H_{12}O_6+C_6H_{12}O_6\)
Mấy pthh sau giống ở trên á bạn
14 tháng 4 2023
(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) \(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
H
12 tháng 4 2023
$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH$
$2CH_3COOH + Mg \to (CH_3COO)_2Mg + H_2$
$(CH_3COO)_2Mg + Ca(OH)_2 \to (CH_3COO)_2Ca + Mg(OH)_2$
$(CH_3COO)_2Ca + K_2CO_3 \to 2CH_3COOK + CaCO_3$
$C_6H_{12}O_6 \xrightarrow{t^o,men\ rượu} 2CO_2 + 2C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH$
$2CH_3COOH + CuO \to (CH_3COO)_2Cu + H_2O$
$(CH_3COO)_2Cu + NaOH \to 2CH_3COONa + Cu(OH)_2$
15 tháng 4 2018
\(2C_4H_{10}+5O_2\underrightarrow{t^o}4CH_3COOH+2H_2O\)
\(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_4đặc]{t^o}CH_3COOC_2H_5+H_2O\)
CH3COOC2H5 + NaOH \(\rightarrow\) CH3COONa + C2H5OH
\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
15 tháng 4 2018
CaC2 + 2H2O\(\rightarrow\) C2H2 + Ca(OH)2
\(C_2H_2+H_2\underrightarrow{t^o}C_2H_4\)
\(C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\)
\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
CO2 + 2NaOH \(\rightarrow\) Na2CO3 + H2O
2CH3COOH + Na2CO3 \(\rightarrow\) 2CH3COONa + CO2 + H2O
TL
13 tháng 4 2023
\(\left(1\right)\) \(CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\)
\(\left(2\right)\) \(C_2H_2+H_2\underrightarrow{Pd,t^o}C_2H_4\)
\(\left(3\right)\) \(C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\)
\(\left(4\right)\) \(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)
\(\left(5\right)\) \(CH_3COOH+C_2H_5OH⇌^{H_2SO_4đ,t^o}CH_3COOC_2H_5+H_2O\)
a, (1) \(C_2H_4+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)
(2) \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
(3) \(2CH_3COOH+Fe\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
b, (1) \(\left(C_6H_{10}O_5\right)_n+nH_2O\underrightarrow{^{H^+}}nC_6H_{12}O_6\)
(2) \(C_6H_{12}O_6\underrightarrow{t^o,xt}2C_2H_5OH+2CO_2\)
(3) \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
(4) \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)