1)
Đốt cháy etilen
\({C_2}{H_4} + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)
Ta có:
\({V_{{O_2}}} = 3{V_{{C_2}{H_4}}} = 11,2.3 = 33,6{\text{ lít}}\)
\( \to {V_{kk}} = 5{V_{{O_2}}} = 33,6.5 = 168{\text{ lít}}\)
2)
Cho hỗn hợp qua brom thì chỉ etilen phản ứng
\({C_2}{H_4} + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)
Ta có:
\({n_{B{r_2}}} = \frac{4}{{80.2}} = 0,025{\text{ mol}} = {n_{{C_2}{H_4}}}\)
\( \to {m_{{C_2}{H_4}}} = 0,025.(12.2 + 4) = 0,7{\text{ gam}}\)