Đáp án:

$\%m_{phenol}=33,82\%\\⇒\%m_{etanol}=66,12\%$

Giải thích các bước giải:

 a. $A+Br_2$:

$C_6H_5OH+3Br_2\to C_6H_2(Br)_3OH↓+3HBr$

$A+Na$

$C_5H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\\0,1\hspace{4cm}0,05\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2$

b. $C_6H_5OH+3Br_2\to C_6H_2(Br)_3OH↓+3HBr$

⇒ $n_↓=\dfrac{19,86}{331}=0,06\ mol⇒n_{phenol}=n_↓ =0,06\ mol⇒ m_{phenol}=94.0,06 =5,64g$

$A+Na$

$C_6H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\\0,06\hspace{4cm}0,03\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2$

$n_{H_2}=\dfrac{3,36}{22,4}=0,15⇔0,03+\dfrac{1}{2}.n_{C_2H_5OH}=0,15 \\⇔ n_{C_2H_5OH} = 0,24\ mol⇒ m_{C_2H_5OH}=0,24.46=11,04g$

⇒ $m_{hh}=11,04+5,64=16,68g\\⇒\%m_{phenol}=\dfrac{5,64}16,68.100\%=33,82\%\\⇒\%m_{etanol}=66,12\%$