Sơ đồ:

\({C_4}{H_{10}}\xrightarrow{{c{\text{r}}acking}}anken + Y\)

Đốt cháy \(C_4H_{10}\)

\(2{C_4}{H_{10}} + 13{{\text{O}}_2}\xrightarrow{{{t^o}}}8C{O_2} + 10{H_2}O\)

\( \to {n_{{O_2}}} = \frac{{13}}{2}{n_{{C_4}{H_{10}}}} = \frac{{13}}{2}.0,15 = 0,975{\text{ mol}}\)

Dẫn sản phẩm cháy qua \(Br_2\) dư thì khí bị giữ lại là anken

\( \to {m_{anken}} = 5,46{\text{ gam}}\)

Anken có dạng \(C_nH_{2n}\) nên có thể quy về \(CH_2\)

\(2C{H_2} + 3{{\text{O}}_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)

\( \to {n_{C{H_2}}} = \frac{{5,46}}{{14}} = 0,39{\text{ mol}} \to {{\text{n}}_{{O_2}}} = 0,39.\frac{3}{2} = 0,585{\text{ mol}}\)

\( \to {n_{{O_2}(Y)}} = 0,975 - 0,585 = 0,39{\text{ mol}}\)

\( \to V = 0,39.22,4 = 8,736{\text{ lít}}\)

Bài Làm:

\(\begin{array}{l}
Bao\,toan\,KL:\\
m_Y=m_{C_4H_{10}}-5,46=0,15.58-5,46=3,24(g)\\
n_Y=n_{C_4H_{10}}=0,15(mol)\\
Dat\,n_{CO_2}=x(mol);n_{H_2O}=y(mol)\\
\to y-x=0,15(1)\\
n_C=n_{CO_2}=x(mol);n_H=2n_{H_2O}=2y(mol)\\
\to 12x+2y=m_Y=3,24(2)\\
(1)(2)\to x=0,21;y=0,36\\
Bao\,toan\,O:\,n_{O_2}=x+\frac{y}{2}=0,39(mol)\\
\to V=0,39.22,4=8,736(l)
\end{array}\)