1)

a)

\(2C{H_4}\xrightarrow{{{{1500}^o}C,lln }}{C_2}{H_2} + 3{H_2}\)

\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)

b)

\(Ca{C_2} + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {C_2}{H_2}\)

\({C_2}{H_2} + {H_2}\xrightarrow{{Pd,PbC{O_3},{t^o}}}{C_2}{H_4}\)

\({C_2}{H_4} + {H_2}\xrightarrow{{Ni.{t^o}}}{C_2}{H_6}\)

c)

Từ \(C_2H_4\) mới ra được chứ em.

\({C_2}{H_4} + {H_2}O\xrightarrow{{{H^ + }}}{C_2}{H_5}OH\)

\({C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}{C_2}{H_4} + {H_2}O\)

\({C_2}{H_4} + B{r_2}\xrightarrow{{}}{C_2}{H_4}B{r_2}\)

2)

Cho hỗn hợp tác dụng với \(Br_2\)

Phản ứng xảy ra:

\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)

Ta có:

\({n_{B{r_2}}} = 0,25.1 = 0,25{\text{ mol}}\)

\( \to {n_{{C_2}{H_2}}} = \frac{1}{2}{n_{B{r_2}}} = 0,125{\text{ mol}}\)

\({n_{C{H_4}}} + {n_{{C_2}{H_2}}} = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol}}\)

\( \to {n_{C{H_4}}} = 0,4 - 0,125 = 0,275{\text{ mol}}\)

\(\% {V_{C{H_4}}} = \frac{{0,275}}{{0,4}} = 68,75\%  \to \% {V_{{C_2}{H_2}}} = 31,25\% \)

\({m_{C{H_4}}} = 0,275.16 = 4,4{\text{ gam;}}{{\text{m}}_{{C_2}{H_2}}} = 0,125.26 = 3,25{\text{ gam}}\)

\( \to \% {m_{C{H_4}}} = \frac{{4,4}}{{4,4 + 3,25}} = 57,5\%  \to \% {m_{{C_2}{H_2}}} = 42,5\% \)