1.
CaC2+ 2H2O -> Ca(OH)2+ C2H2
C2H2+ H2 $\buildrel{{Pd, t^o}}\over\longrightarrow$ C2H4
nCH2=CH2 $\buildrel{{t^o, p, xt}}\over\longrightarrow$ (CH2-CH2)n
C2H2+ HCl $\buildrel{{xt, t^o}}\over\longrightarrow$ CH2=CHCl
nCH2=CHCl $\buildrel{{t^o, p, xt}}\over\longrightarrow$ (CH2-CHCl)n
C2H4+ H2O $\buildrel{{H+, t^o}}\over\longrightarrow$ C2H5OH
C2H5OH+ O2 $\buildrel{{\text{lên men giấm}}}\over\longrightarrow$ CH3COOH+ H2O
CH3COOH+ C2H5OH $\buildrel{{H2SO4, t^o}}\over\rightleftharpoons$ CH3COOC2H5+ H2O
nC2H2 $\buildrel{{CuCl2, t^o}}\over\longrightarrow$ C4H4
C4H4+ H2 $\buildrel{{Pb, t^o}}\over\longrightarrow$ CH2=CH-CH=CH2
nCH2=CH-CH=CH2 $\buildrel{{t^o, p, xt}}\over\longrightarrow$ (CH2-CH=CH-CH2)n
2.
a,
nCa(HCO3)2= 0,1 mol
CaCO3+ H2O+ CO2 -> Ca(HCO3)2
=> nCO2= nCaCO3 tan= 0,1 mol
nCaCO3 ko tan= 0,2 mol
=> Có 0,2+0,1= 0,3 mol CaCO3 tạo thành
CO2+ Ca(OH)2 -> CaCO3+ H2O
=> nCO2= 0,3 mol
Tổng nCO2= 0,4 mol
=> mCO2= 17,6g
mdd tăng= mCO2+ mH2O - mCaCO3 tách ra
=> mH2O= 7,2g
=> nH2O= 0,4 mol
=> nH= 2nH2O= 0,8 mol
=> mH= 0,8g
nC= nCO2 => mC= 4,8g
=> mO= 12-4,8-0,8= 6,4g
=> nO= 0,4 mol
nC: nH: nO= 1:2:1
=> CTĐGN (CH2O)n
M= 3,75.16= 60
=> n=2 => CTPT C2H4O2
b,
A làm quỳ hoá đỏ => A có nhóm chức axit
=> CTCT là CH3COOH
CaCO3+ 2CH3COOH -> (CH3COO)2Ca+ CO2+ H2O
KOH+ CH3COOH -> CH3COOK+ H2O
CH3COOH+ Na -> CH3COONa+ 1/2H2
BaO+ 2CH3COOH -> (CH3COO)2Ba+ H2O
CH3COOH+ C2H5OH $\buildrel{{t^p, H2SO4}}\over\rightleftharpoons$ CH3COOC2H5+ H2O
