Sơ đồ phản ứng:
\(C_4H_{10}\underrightarrow{^{cracking}}ankan+anken\)
Ta có:
\(n_X=\frac{8,96}{22,4}=0,4\left(mol\right)\)
Cho hỗn hợp X qua brom thì có 25,6 gam brom phản ứng.
\(n_{Br2}=n_{anken}=\frac{25,6}{80.2}=0,16\left(mol\right)\)
\(\Rightarrow n_{ankan}=n_{C4H10\left(bđ\right)}=0,4-0,16=0,24\left(mol\right)\)
BTKL:
\(m_X=m_{C4H10}=0,24.\left(12.4+10\right)=13,92\left(g\right)\)
\(\Rightarrow M_X=\frac{13,92}{0,4}=34,8\)
\(\Rightarrow D=d_{X/H2}=\frac{34,8}{2}=17,4\)