Đáp án: metanol
Giải thích các bước giải:
$2ROH+2Na\to 2RONa+H_2$
$n_{RONa}=\dfrac{4,32}{M_R+39}(mol)$
$\to n_{ROH}=n_{RONa}=\dfrac{4,32}{M_R+39}(mol)$, $n_{H_2}=\dfrac{n_{RONa}}{2}=\dfrac{2,16}{M_R+39}(mol)$
$\Delta m=2,48g\to \dfrac{4,32.(M_R+17)}{M_R+39}-\dfrac{2.2,16}{M_R+39}=2,48$
$\to 2,48(M_R+39)=4,32(M_R+17)-4,32$
$\to M_R=15(CH_3)$
Vậy $T: CH_3OH$ (metanol)