m(C6H10O5)n=40,5.\(\dfrac{80}{100}\)=32,4kg

→n(C6H10O5)n=\(\dfrac{32,4}{162}\)=0,2kmol

→nethanol=0,2.2.\(\dfrac{78}{100}\)0,312kmol

→V=\(\dfrac{ }{ }\)m/D=\(\dfrac{0,312.10^3.64}{0,8}\)=17940

→malkene=0,8gam

→malkane=2-0,8=1,2gam

→%Alkene=\(\dfrac{0,8}{2}\).100=40%

→%Alkane=\(\dfrac{1,2}{2}\).100=60%

(1)CaC2+2H2O→C2H2+Ca(OH)2

(2)C2H2+H2→C2H4

(3)C2H4+H2O→C2H5OH

(4).C2H5OH+CuO→CH3CHO+Cu+H2O

nC2H5Cl=\(\dfrac{25,8}{64,5}\)=0,4mol

→nC2H4=0,4.\(\dfrac{100}{80}\)=0,5mol

→VC2H4=0,5.24,79=12,395L

nH2O=\(\dfrac{7,56}{18}\)=0,42(mol);nCO2=\(\dfrac{14,08}{44}\)=0,32(mol)

Có :nH2O>nCO2(0,42>0,32)

=>Hydrocarbon:ankan

nhh.ankan=nH2O-nCO2=0,1(mol)

3<số .C.trung.bình =\(\dfrac{ }{ }\)nCO2/nhh=\(\dfrac{0,32}{0,1}\)3,2<4

(1)CH3-CH2-CH2Cl+HCl

(2)CH3-CH3

(3)CH3-CHBr-CH3

(4)C6H6Cl6

(5)CH3-CO-CH3+H2O+Cu