\(\begin{array}{l} {C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\ {C_3}{H_4} + 2B{r_2} \to {C_3}{H_4}B{r_4}\\ {C_3}{H_4} + AgN{O_3} + N{H_3} \to {C_3}{H_3}Ag + N{H_4}N{O_3}\\ V{C_2}{H_6} = 1,344l\\ n{C_3}{H_3}Ag = \dfrac{{13,23}}{{147}} = 0,09\,mol\\ n{C_3}{H_4} = n{C_3}{H_3}Ag = 0,09\,mol\\ \% V{C_2}{H_6} = \dfrac{{1,344}}{{6,72}} \times 100\%  = 20\% \\ \% V{C_3}{H_4} = \dfrac{{0,09 \times 22,4}}{{6,72}} \times 100\%  = 30\% \\ \% V{C_2}{H_4} = 100 - 20 - 30 = 50\% \\ b)\\ 2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O\\ {C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\ {C_3}{H_4} + 4{O_2} \to 3C{O_2} + 2{H_2}O\\ n{O_2} = 0,06 \times 3,5 + 0,15 \times 3 + 0,09 \times 4 = 1,02mol\\ Vkk = 1,02 \times 5 \times 22,4 = 114,24l \end{array}\)