Đáp án A
$0,06\left\{ \begin{array}{l} X:{C_n}{H_{2n}}:x\\ Y,Z:{C_m}{H_{2m + 2 - 2k}}:y \end{array} \right.$ $\xrightarrow{{{O}_{2}}:0,215\text{ mol}}$$\left\{ \begin{array}{l} C{O_2}:0,15\\ {H_2}O \end{array} \right.$
BT O: $2{{n}_{{{O}_{2}}}}=2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}\to {{n}_{{{H}_{2}}O}}=0,13\text{ mol}$
${{\text{n}}_{Y,Z}}=\frac{{{n}_{C{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}}{k-1}\to y\left( k-1 \right)=0,02$
TH: $k=2\to y=0,02,\text{ }x=0,04$
BT C: $0,04.n+0,02.m=0,15$
Với $n=2\to m=3,5$
$\begin{array}{l} \to 0,06\left\{ \begin{array}{l} X:{C_2}{H_4}:0,04\\ {C_3}{H_4}:a\\ {C_4}{H_6}:b \end{array} \right.\\ \to \left\{ \begin{array}{l} a + b = 0,02\\ 3a + 4b = 0,15 - 0,04.2 \end{array} \right.\\ \to \left\{ \begin{array}{l} a = 0,01\\ b = 0,01 \end{array} \right. \end{array}$
$%{{m}_{{{C}_{3}}{{H}_{4}}}}=19,42%$
