\(m(C6H10O5)n=40,5.80100=32,4\) kg​

⇒ \(n(C6H10O5)n=32,4162=0,2\) kmol

 ⇒ \(nethanol=0,2.2.78100=0,312\) kmol

⇒ \(V=mD=0,312.103.460,8=17940\) mL

=> 17,94L

=>  \(Malkene=0,8\) gam

=>  \(Malkane=2-0,8=1,2\) gam.

=> %alkane = \(\frac{0,8}{2}\) x 100 = 40%

=> %alkane = \(\frac{1,2}{2}\) x100 = 60%\(%Alkene=0,82.100=40%%Alkene=20,8​.100=40%\)

⇒ \(%Alkane=1,22.100=60%%Alkane=21,2​.100=60%\)\(%Alkane=21,2​.100=60%\)\(%Alkane=21,2​.100=60%\)

(1) \(CaC2+2H2O→C2H2+Ca(OH)2CaC2​+2H2​O→C2​H2​+Ca(OH)2​\)

(2) \(C2H2+H2\underset{}{\overset{\text{Pd/PdCO 3 ​ }}{\xrightarrow{}}}C2H4C2H2+H2\to Pd/PdCO3C2H4\)

(3) \(C2H4+H2O\underset{}{\overset{H3PO4.t^{\omicron}\text{ }}{\xrightarrow[.]{}}}C2H5OHC2H4+H2O\to H3PO4.toC2H5OH\)

(4)\(C2H5OH+CuO\underset{}{\overset{t^{\omicron}}{\xrightarrow{}}}CH3CHO+Cu+H2O\)

PTHH:

\(\left(C H\right)_{2} = \left(C H\right)_{2} + H C l \rightarrow \left(C H\right)_{3} \left(C H\right)_{2} C l\)

Ta có:

\(nC2H5Cl=25,864,5=0,4nC2​H5​Cl​=64,525,8​=0,4\) mol

→ \(nC2H4=0,4.10080=0,5nC2​H4​​=0,4.80100​=0,5\) mol.

→ \(VC2H4=0,5.24,79=12,395VC2H4=0,5.24,79=12,395\) 

\(nH2​O​=187,56​=0,42\) mol

\(nCO2=14,0844=0,32nCO2​​=4414,08​=0,32\) mol

ta co cong thuc:  \(Cn​H2n+2​\).

PTHH:

Cn​H2n+2​+23n+1​O2​→tonCO2​+(n+1)H2​O

=> \(nalkane=nH2O-nCO2=0,42-0,32=0,1\) mol

=> \(\overline{n}=nCO2nalkane=0,320,1=3,2\)

1. -> \(\begin{cases}CH3-CHCL-CH3\\ CH2CL-CH2-CH3\end{cases}\) + HCL

2. -> CH3-CH3

3. -> \(\begin{cases}CH3-CHBr-CH3\\ CH2Br-CH2-CH3\end{cases}\)

4. -> C6H6Cl

5. ->  CH3-CO-CH3+H20+Cu