\(CT:C_nH_{2n+2}\left(1mol\right)\)
\(C_nH_{2n+2}\underrightarrow{t^0,xt}C_nH_{2n}+H_2\)
\(1.......................1...........1\)
\(\overline{M}=\dfrac{14n+2}{2}=7.5\cdot2=15\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow n=2\)
\(\%C=\dfrac{12\cdot2}{12\cdot2+6}\cdot100\%=80\%\)
Ta có: \(\dfrac{n_t}{n_s}=\dfrac{M_s}{M_t}\Rightarrow\dfrac{1}{2}=\dfrac{18.2}{M_t}\Rightarrow M_t=72\) (Do cracking hoàn toàn)
Do đó X là $C_5H_{12}$. Không phân nhánh vậy X là pentan
\(CT:C_nH_{2n+2}\left(1mol\right)\)
\(C_nH_{2n+2}\underrightarrow{t^0,xt}C_aH_{2a+2}+C_bH_{2b}\left(n=a+b\right)\)
\(1...........................1.............1\)
\(m_Y=2\cdot2\cdot18=72\left(g\right)\)
\(BTKL:\)
\(m_X=m_Y=72\left(g\right)\)
\(\Rightarrow M_X=14n+2=72\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow n=5\)
\(heptan\)