a,
Theo lí thuyết: $m_{\text{anđehit}}= 1,16:80\%= 1,45g$
$RCH_2OH + \frac{1}{2}O_2\to RCHO+ H_2O$
$n_{\text{ancol}}=n_{\text{anđehit}}$
$\Rightarrow \frac{1,5}{R+14+17}=\frac{1,45}{R+29}$
$\Leftrightarrow 1,45(R+31)=1,5(R+29)$
$\Leftrightarrow R=29 (C_2H_5)$
Vậy ancol là $C_2H_5OH$ hay $C_2H_6O$
b,
$C_4H_{10}\buildrel{{\text{cracking}}}\over\to C_2H_4+C2H_6$
$C_2H_4+H_2O\buildrel{{H^+, t^o}}\over\to C_2H_5OH$