\(\begin{array}{l} 3)\\ {C_n}{H_{2n}} + B{r_2} \to {C_n}{H_{2n}}B{r_2}\\ n{C_n}{H_{2n}} = nB{r_2} = \frac{{40}}{{160}} = 0,25\,mol\\  =  > M{C_n}{H_{2n}} = \frac{{9,1}}{{0,25}} = 36,4\\  =  > 14n = 36,4 =  > n = 2,6\\  =  > {C_2}{H_4},{C_3}{H_6}\\ b)\\ {C_2}{H_4}(a\,mol),{C_3}{H_6}(b\,mol)\\ a + b = 0,25\\ 28a + 42b = 9,1\\  =  > a = 0,1b = 0,15\\ \% m{C_2}{H_4} = \frac{{0,1 \times 28}}{{9,1}} \times 100\%  = 30,77\% \\ \% m{C_3}{H_6} = 100 - 30,77 = 69,23\% \\ 4)\\ a)\\ nC{O_2} = \frac{{24,64}}{{22,4}} = 1,1\,mol =  > nC = 1,1\,mol\\ n{H_2}O = nC{O_2} = 1,1 =  > nH = 2,2\,mol\\ m = 1,1 \times 12 + 2,2 = 15,4g\\ b)\\ BTNT\,O:2n{O_2} = 2nC{O_2} + n{H_2}O\\  =  > n{O_2} = 1,65\,mol\\ V = 1,65 \times 22,4 = 36,96l \end{array}\)