Vì hỗn hợp Y làm nhạt màu dung dịch KMnO4 nên Y có anken dư và H2 hết

\(\begin{array}{l} 0,35\,\,mol\,\,X\left\{ \begin{array}{l} {H_2}\\ {C_x}{H_{2x}}\\ {C_y}{H_{2y + 2}} \end{array} \right. \to 0,3\,\,mol\,Y\left\{ \begin{array}{l} {C_x}{H_{2x}}\\ {C_x}{H_{2x + 2}}\\ {C_y}{H_{2y + 2}} \end{array} \right. \to 0,2\,\,mol\,\,Z\left\{ \begin{array}{l} {C_x}{H_{2x + 2}}\\ {C_y}{H_{2y + 2}} \end{array} \right.\\ = > \,{n_{{H_2}}} = {n_X} - {n_Y} = 0,05\,mol\\ \left\{ \begin{array}{l} {n_{{C_x}{H_{2x}}\,(Y)\,\,}} = \,{n_Y} - {n_Z} = 0,1\\ {m_{{C_x}{H_{2x}}(Y)}} = 2,8 \end{array} \right.\left. \begin{array}{l} \\ \end{array} \right\} = > {M_{{C_x}{H_{2x}}}} = 28 = > {C_x}{H_{2x}}\,\,\,({C_2}{H_4})\\ {n_{{C_x}{H_{2x + 2}}}} = \,{n_{{H_{2(p.u)}}}} = 0,05 = > {n_{{C_y}{H_{2y + 2}}}} = 0,2 - 0,05 = 0,15;\,\,{m_Z} = 0,2.40,5 = 0,05.30 + 0,15(14y + 2)\\ = > y = \,3\,\,({C_3}{H_8})\\ = > \% V\,cac\,khi\,trong\,Y\, \end{array}\)