a, Ta có:
\(\left\{{}\begin{matrix}n_{CO2}=0,2\left(mol\right)\\n_{H2O}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H2O}-n_{CO2}=0,3-0,2=0,1\left(mol\right)\)
Suy ra A là ankan
\(\overline{C}=\frac{n_{CO2}}{n_{ankan}}=2\Rightarrow C_2H_4\)
b, PTHH:
\(C_2H_4+Cl_2\rightarrow C_2H_3Cl+HCl\)
\(C_2H_4+Cl_2\rightarrow C_2H_4Cl_2\)
Giải hệ PT:
\(\left\{{}\begin{matrix}a+b=0,0625\\62,5a+99b=2,475\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,004\\b=0,022\end{matrix}\right.\)
C2H3Cl: monome vinyl clorua
C2H4Cl2: 1,2-dichloroethane