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\(2C_4H_{10}+5O_2\underrightarrow{t^o}4CH_3COOH+2H_2O\)
\(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_4đặc]{t^o}CH_3COOC_2H_5+H_2O\)
CH3COOC2H5 + NaOH \(\rightarrow\) CH3COONa + C2H5OH
\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
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
Câu 2:
\(1,C_{12}H_{22}O_{11}+H_2O\underrightarrow{t^o}C_6H_{12}O_6+C_6H_{12}O_6\\ C_6H_{12}O_6\xrightarrow[\text{men rượu}]{H^+,t^o}2C_2H_5OH+2CO_2\uparrow\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ 2CH_3COOH+ZnO\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)
2, tất cả đều giống 1 trừ PTHH cuối:
\(2CH_3COOH+MgO\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)
Câu 3:
Cho mẩu Na tác dụng với từng chất:
- Na tan dần, có sủi bọt khí không màu, mùi: C2H5OH, CH3COOH (*)
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\\ CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\uparrow\)
- Không hiện tượng: C6H12O6
Cho QT vào các chất (*):
- Hoá hồng: CH3COOH
- Không hiện tượng: C2H5OH
Câu 4:
a) Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{44}{44}=1\left(mol\right)\)
Bảo toàn H: \(n_H=2n_{H_2O}=2.\dfrac{27}{18}=3\left(mol\right)\)
Xét mH + mC = 3 + 12 = 15 (g)
=> A chỉ chứa C và H
b) MA = 2.15 = 30 (g/mol)
CTPT: CxHy
=> x : y = 1 : 3
=> (CH3)n = 30
=> n = 2
CTCT: CH3-CH3
Câu 5:
a) Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{44}{44}=1\left(mol\right)\)
Bảo toàn H: \(n_H=2n_{H_2O}=2.\dfrac{18}{18}=2\left(mol\right)\)
Xét mH + mC = 2 + 12 = 14 (g)
=> A chỉ chứa C và H
b) MA = 2.14 = 28 (g/mol)
CTPT: CxHy
=> x : y = 1 : 2
=> (CH2)n = 28
=> n = 2
CTPT: C2H4
CTCT: CH2=CH2
Câu 6:
\(a,n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{46}{46}=1\left(mol\right)\)
PTHH: CH3COOH + C2H5OH -H2SO4 (đặc), to-> CH3COOC2H5 + H2O
LTL: 0,5 < 1 => C2H5OH dư
Theo pthh: nCH3COOC2H5 = nCH3COOH = 0,5 (mol)
=> mCH3COOC2H5 = 0,5.60%.88 = 26,4 (g)
Câu 7:
\(a,n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\)
PTHH:
CH3COOH + C2H5OH -H2SO4 (đặc), to-> CH3COOC2H5 + H2O
LTL: 1 > 0,5 => CH3COOH dư
Theo pthh: nCH3COOC2H5 = nC2H5OH = 0,5 (mol)
=> mCH3COOC2H5 = 0,5.70%.88 = 30,8 (g)
C2H6 \(\underrightarrow{t^o}\) C2H4
C2H4 + H2O \(\underrightarrow{axit}\) C2H5OH
C2H5OH + O2 \(\underrightarrow{mengiam}\)CH3COOH
CH3COOH + C2H5OH → CH3COOC2H5 + H2O
NaOH + CH3COOC2H5 → C2H5OH + CH3COONa
A/ C6H12O6 => (men rượu,to) 2C2H5OH + 2CO2
C2H5OH + O2 => (men giấm) CH3COOH + H2O
2CH3COOH + CaCO3 => (CH3COO)2Ca + CO2 + H2O
(CH3COO)2Ca + H2SO4 => CaSO4 + 2CH3COOH
B/ NaOH + CH3COOH =>(to) CH3COONa + H2O
CH3COONa + HCl => (to) CH3COOH + NaCl
CH3COOH + C2H5OH => (pứ hai chiều, xt:H2SO4đ,to) CH3COOC2H5 + H2O
CH3COOC2H5 + H2O =>(pứ hai chiều, xt: to,H2SO4đ) CH3COOH + C2H5OH
A) C6H12O6----->2C2H5OH+2CO2 (BAY LÊN)
C2H5OH+O2----->CH3COOH+H2O
CH3COOH+Ca(OH)2----->(CH3COO)2Ca+2H2O
(CH3COO)2Ca+H2SO4----->CaSO4+2CH3COOH
TÔI GIẢI CHO CẬU RỒI ĐÓ
CỐ LÊN NHA ^^
A/ C6H12O6 => (men rượu,to) 2C2H5OH + 2CO2
C2H5OH + O2 => (men giấm) CH3COOH + H2O
2CH3COOH + CaCO3 => (CH3COO)2Ca + CO2 + H2O
(CH3COO)2Ca + H2SO4 => CaSO4 + 2CH3COOH
B/ NaOH + CH3COOH => CH3COONa + H2O
CH3COONa + HCl => CH3COOH + NaCl
CH3COOH + C2H5OH => (pứ hai chiều, xt:H2SO4đ,to) CH3COOC2H5 + H2O
CH3COOC2H5 + H2O => CH3COOH + C2H5OH (pứ hai chiều, xt: to,H2SO4đ)
C2H4 + H2O \(\underrightarrow{lênmen}\) C2H5OH
C2H5OH + O2 \(\underrightarrow{lênmen}\) CH3COOH + H2O
CH3COOH + C2H5OH \(\rightarrow\) CH3COOC2H5 + H2O
CH3COOC2H5 + NaOH \(\rightarrow\) CH3COONa + C2H5OH
Chuỗi 1:
\(\left(1\right)CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ \left(2\right)CaO+3C\rightarrow\left(2000^oC,lò.điện\right)CaC_2+CO\uparrow\\ \left(3\right)CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\\ \left(4\right)C_2H_2+H_2\rightarrow\left(Ni,t^o\right)C_2H_4\\ \left(5\right)C_2H_4+H_2O\rightarrow\left(t^o,H^+\right)C_2H_5OH\\ \left(6\right)C_2H_5OH+2NaOH+CH_3COOH\rightarrow CH_3COONa+C_2H_5ONa+2H_2O\)
$C + O_2 \xrightarrow{t^o} CO_2$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$CO_2 + C \to 2CO$
$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + NaOH \to CH_3COONa + C_2H_5OH$