\(\overline{M_A}=8,1.2=16,2\)
\(C_nH_{2n-2}+2H_2\rightarrow C_2H_{2n+2}\)
\(A\left\{{}\begin{matrix}n_{H2}:a\left(mol\right)\\n_{ankin}:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow B\left\{{}\begin{matrix}n_{ankan}:b\left(mol\right)\\n_{H2}:a-2b\left(mol\right)\end{matrix}\right.\)
Giả sử \(m_A=m_B=16,2\)
\(\overline{M_B}=40,5\)
\(\Rightarrow n_A=1\left(mol\right)\)
\(n_B=0,4\left(mol\right)\)
\(\left\{{}\begin{matrix}a+b=1\\a+b-2b=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,7\\b=0,3\end{matrix}\right.\)
Ta có:
\(m_A=m_{H2}+m_{ankin}\Leftrightarrow16,2=0,7.2+0,3.\left(14n-2\right)\)
\(\Leftrightarrow n=3,6\left(C_3H_4;C_4H_6\right)\)
\(\left\{{}\begin{matrix}n_{C3H4}:x\left(mol\right)\\n_{C4H6};y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=0,3\\40x+54y=16,2-0,7.2=14,8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%_{H2}=\frac{0,7.100}{1}=70\%\)
\(\%_{C3H4}=10\%\Rightarrow\%_{C3H6}=20\%\)
Trong B: \(\left\{{}\begin{matrix}n_{H2}=0,1\left(mol\right)\\n_{C3H8}=0,1\left(mol\right)\\n_{C4H10}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%_{H2}=25\%=\%_{C3H8}\)
\(\Rightarrow\%_{C4H10}=50\%\)
