$Etanol:C_{2}H_{5}OH$
$Phenol:C_{6}H_{5}OH$
Gọi số mol $C_{2}H_{5}OH$: y
$C_{6}H_{5}OH$:y
$C_{2}H_{5}OH+Na \to C_{2}H_{5}ONa+1/2H_{2}$
$C_{6}H_{5}OH +Na \to C_{6}H_{5}ONa+1/2H_{2}$
$mC_{2}H_{5}OH+mC_{6}H_{5}OH=18,6$
⇒$46x+94y=18,6(1)$
$nH_{2}=\frac{3,36}{22,4}=0,15$
⇒$1/2x+1/2y=0,15(2)$
(1)(2)⇒$\left \{ {{x=0,2} \atop {y=0,1}} \right.$
$\%mC_{2}H_{5}OH=\frac{0,2.46}{18,6}.100=49,46\%$
$\%mC_{6}H_{5}OH=100-49,46=50,54\%$