Đáp án:
23%,34,48%,42,52%
Giải thích các bước giải:
\(\begin{array}{l} 2AgN{O_3} + {C_2}{H_2} + 2N{H_3} \to 2N{H_4}N{O_3} + A{g_2}{C_2}\\ {C_3}{H_6} + B{r_2} \to {C_3}{H_6}B{r_2}\\ nA{g_2}{C_2} = \frac{{88,8}}{{240}} = 0,37mol\\ = > n{C_2}{H_2} = 0,37mol\\ n{C_3}{H_6} = \frac{{12,6}}{{42}} = 0,3mol\\ nhh = \frac{{19,488}}{{22,4}} = 0,87mol\\ = > nC{H_4} = 0,2mol\\ = > \% VC{H_4} = 23\% \\ \% V{C_3}{H_6} = 34,48\% \\ \% V{C_2}{H_2} = 42,52\% \end{array}\)