Ta có:
\(n_{CO2}=n_C=\frac{8,8}{44}=0,2\left(mol\right)\)
\(\Rightarrow m_C=0,2.12=2,4\left(g\right)\)
\(\Rightarrow n_H=2n_{H2O}=\frac{4,5.2}{18}=0,5\left(mol\right)\)
\(\Rightarrow m_H=0,5.1=0,5\left(g\right)\)
\(m_A=m_C+m_H\)
=> A là hidrocacbon
\(n_C:n_H=2:5\)
Nên CTĐGN :(C2H5)n
\(n_B=\frac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow M_B=\frac{5,8}{0,1}=58\)
\(\Leftrightarrow n=2\left(C_4H_{10}\right)\)
CTCT:
(A): CH3 − CH2 − CH2 − CH3
(B): CH3 − CH(CH3) − CH3