Ta có: \(\left\{{}\begin{matrix}M_X=21,3.2=42,6\left(g/mol\right)\\M_Y=14,2.2=28,4\left(g/mol\right)\end{matrix}\right.\)
Quy đổi hỗn hợp \(\left\{{}\begin{matrix}C_2H_6\\C_3H_8\end{matrix}\right.\) về \(C_nH_{2n+2}\)
Giả sử có 1 mol X \(\Rightarrow m_X=1.42,6\left(g\right)\)
Gọi \(n_{X\left(pư\right)}=x\left(mol\right)\left(0< x\le1\right)\)
PTHH: \(C_nH_{2n+2}\xrightarrow[xt]{t^o}C_nH_{2n}+H_2\)
x------------>x------->x
`=>` \(Y\left\{{}\begin{matrix}C_nH_{2n+2}:1-x\left(mol\right)\\C_nH_{2n}:x\left(mol\right)\\H_2:x\left(mol\right)\end{matrix}\right.\)
`=>` \(n_Y=1-x+x+x=x+1\left(mol\right)\)
Theo ĐLBTKL: \(m_Y=m_X=42,6\left(g\right)\)
`=>` \(M_Y=\dfrac{42,6}{x+1}=28,4\left(g/mol\right)\)
`=>` \(x=0,5\left(mol\right)\)
`=>` \(H=\dfrac{0,5}{1}.100\%=50\%\)