Các anken đều có dạng \({C_n}{H_{2n}}{\text{ (n}} \geqslant {\text{2)}}\)

a) \({C_n}{H_{2n}}{\text{  + B}}{{\text{r}}_2}\xrightarrow{{}}{C_n}{H_{2n}}B{r_2}\)

Ta có: 

\({m_{B{r_2}}} = 150.40\%  = 60{\text{ gam}} \to {{\text{n}}_{B{r_2}}} = {n_X} = \frac{{60}}{{80.2}} = 0,375{\text{ mol}} \to {{\text{M}}_X} = 14n = \frac{{10,5}}{{0,375}} = 28 \to n = 2\)

Vậy X là \({C_2}{H_4}\)

CTCT: \(C{H_2} = C{H_2}\)

b) \({C_n}{H_{2n}} + {H_2}\xrightarrow{{Ni,{t^o}}}{C_n}{H_{2n + 2}}\)

Áp dụng công thức: \(pV = nRT\)

Với p=1 atm; V=0,4928 lít; R=0,082; R=27,3+273=300,3K

\(\to {n_{{H_2}}} = \frac{{1.0,4928}}{{0,082.300,3}} = 0,02{\text{ mol = }}{{\text{n}}_Y} \to {M_Y} = 14n = \frac{{1,4}}{{0,02}} = 70 \to n = 5\)

Vậy Y là \({C_5}{H_{10}}\)

CTCT:

\(C{H_2} = CH - C{H_2} - C{H_2} - C{H_3};{\text{ C}}{{\text{H}}_3} - CH = CH - C{H_2} - C{H_3}\)

\(C{H_2} = C(C{H_3}) - C{H_2} - C{H_3};{\text{ C}}{{\text{H}}_3} - C(C{H_3}) = CH - C{H_3};{\text{ C}}{{\text{H}}_3} - CH(C{H_3}) - CH = C{H_2}\)