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

Đáp án A

$0,06\left\{\begin{array}{l}X:C_n{H}_{2n}:x\\ Y,Z:C_m{H}_{2m+2-2k}:y\end{array}\right.\underset{}{\overset{O_2:0,215\text{ mol}}{\to }}\left\{\begin{array}{l}C{O}_2:0,15\\ H_{2}O\end{array}\right.$

BT O: $2n_{O_2}=2n_{C{O}_2}+n_{H_{2}O}\to n_{H_{2}O}=0,13\text{ mol}$

${\text{n}}_{Y,Z}=\frac{n_{C{O}_2}-n_{H_{2}O}}{k-1}\to y\left(k-1\right)=0,02$

TH: $k=2\to y=0,02,x=0,04$

BT C: $0,04.n+0,02.m=0,15$

Với $n=2\to m=3,5$

$$\begin{gathered} \to 0,06\left\{\begin{array}{l}X:C_2{H}_4:0,04\\ C_3{H}_4:a\\ C_4{H}_6:b\end{array}\right.\to \left\{\begin{array}{l}a+b=0,02\\ 3a+4b=0,15-0,04.2\end{array}\right.\to \left\{\begin{array}{l}a=0,01\\ b=0,01\end{array}\right. \\ \%m_{C_3{H}_4}=19,42\% \end{gathered}$$

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