Theo ĐLBT KL, có: mancol + mNa = mcr + mH2
⇒ mH2 = 12,4 + 9,2 - 21,3 = 0,3 (g) \(\Rightarrow n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\)
\(ROH+Na\rightarrow RONa+\dfrac{1}{2}H_2\)
\(n_{ROH}=2n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow\overline{M}_{ancol}=\dfrac{12,4}{0,3}=\dfrac{124}{3}\left(g/mol\right)\)
\(\Rightarrow M_R=24,3\left(g/mol\right)\)
→ CH3OH và C2H5OH