\(C_nH_{2n}+Br_2\rightarrow C_nH_{2n-2}Br_4\)
\(n_{Br_2}=\frac{3,2}{160}=0,2\left(mol\right)=n_{anken}\)
\(M_{ankin}=\frac{8,4}{0,2}=42\left(g\right)\)
anken | \(C_2H_4\) | \(C_3H_6\) |
M | 28 | 42 |
vậy anken đó là:\(C_3H_6\)
sgk trang 110 nha bạn
4/
%\(_{M_H}\)=100-83,72=16,28%
ta có:
\(\%C:\%H=\frac{83,72}{12}16,28\approx\frac{3}{7}\)
\(\Rightarrow CTĐGN\) là :\(\left(C_3H_7\right)n\)
n=1 | n=2 |
\(C_3H_7\) | \(C_6H_{14}\) |
\(C_nH_{2n+1}\) | \(C_nH_{2n+2}\) |
vây X là:\(C_6H_{14}\)