Hai ankanol : CnH2n+1OH
\(n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ 2C_nH_{2n+1}OH + 2Na \to 2C_nH_{2n+1}ONa + H_2\\ n_{C_nH_{2n+1}ONa} = 2n_{H_2} = 0,3(mol)\\ \Rightarrow 0,3(14n + 40) = 17,6\\ \Rightarrow n = 1,33\)
Vậy hai ancol là : \(CH_3OH,C_2H_5OH\)
\(n_{Na} = 2n_{H_2} = 0,3(mol)\\ m_{ankanol} = m_{hh\ rắn} + m_{H_2} - m_{Na} = 17,6 + 0,15.2 - 0,3.23 = 11(gam)\)