Đốt cháy hoàn toàn 0,06 mol hỗn hợp gồm 1 anken X và 2 hiđrocacbon...

$0,06\left\{ \begin{aligned}
& X:{{C}_{n}}{{H}_{2n}}:x \\
& Y,Z:{{C}_{m}}{{H}_{2m+2-2k}}:y \\
\end{aligned} \right.\xrightarrow[{}]{{{O}_{2}}:0,215\text{ mol}}\left\{ \begin{aligned}
& C{{O}_{2}}:0,15 \\
& {{H}_{2}}O \\
\end{aligned} \right.$
BT O: $2{{n}_{{{O}_{2}}}}=2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}\to {{n}_{{{H}_{2}}O}}=0,13\text{ mol}$
${{\text{n}}_{Y,Z}}=\dfrac{{{n}_{C{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}}{k-1}\to y\left( k-1 \right)=0,02$
TH: $k=2\to y=0,02,x=0,04$
BT C: $0,04.n+0,02.m=0,15$
Với $n=2\to m=3,5$
$\to 0,06\left\{ \begin{aligned}
& X:{{C}_{2}}{{H}_{4}}:0,04 \\
& {{C}_{3}}{{H}_{4}}:a \\
& {{C}_{4}}{{H}_{6}}:b \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a+b=0,02 \\
& 3a+4b=0,15-0,04.2 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,01 \\
& b=0,01 \\
\end{aligned} \right.$
$\%{{m}_{{{C}_{3}}{{H}_{4}}}}=19,42\%$.