Đáp án:
$m_{phenol}=9,4g\\m_{etanol}=13,8g$
Giải thích các bước giải:
$A+Br_2$:
$C_6H_5OH+3Br_2\to C_6H_2(Br)_3OH↓+3HBr$
⇒ $n_↓=\dfrac{33,1}{331}=0,1\ mol⇒n_{phenol}=n_↓ =0,1\ mol⇒ m_{phenol}=94.0,1 =9,4g$
$A+Na$
$C_5H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\\0,1\hspace{4cm}0,05\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2⇔0,05+\dfrac{1}{2}.n_{C_2H_5OH}=0,2 \\⇔ n_{C_2H_5OH} = 0,3\ mol⇒ m_{C_2H_5OH}=0,3.46=13,8g$