4)

Phản ứng xảy ra:

\({C_2}{H_4} + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)

Ta có:

\({n_{{C_2}{H_4}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)

Theo phản ứng:

\({n_{C{O_2}}} = 2{n_{{C_2}{H_4}}} = 0,3.2 = 0,6{\text{ mol}}\)

\( \to {m_{C{O_2}}} = 0,6.44 = 26,4{\text{ gam}}\)

Dẫn \(CO_2\) vào \(Ca(OH)_2\) dư

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

Ta có:

\({n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,6{\text{ mol}}\)

\( \to {m_{CaC{O_3}}} = 0,6.100 = 60{\text{ gam}}\)

5)

Phản ứng xảy ra:

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

Ta có:

\({n_{B{r_2}}} = \frac{8}{{80.2}} = 0,05{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}}\)

\( \to {V_{{C_2}{H_4}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

\( \to \% {V_{{C_2}{H_4}}} = \frac{{1,12}}{{3,36}} = 33,33\%  \to \% {V_{C{H_4}}} = 66,67\% \)