Đáp án:
a/ $\%m_{phenol}=50,54\%⇒\%m_{ancol}=49,46\%$
b/ 2,24 lít
Giải thích các bước giải:
$C_6H_5OH+3HNO_3\xrightarrow{H_2SO_4đ}(O_2N)_3C_6H_2OH↓+3H_2O$
$n_↓ =\dfrac{45,8}{229}=0,2\ mol$
Có: $n_{phenol}=n_↓=0,2 ⇒m_{phenol}=0,2.94=18,8g ⇒ m_{C_2H_5OH}=37,2-18,8=18,4g$
Vậy: $\%m_{phenol}=\dfrac{18,8}{37,2}.100\%=50,54\%⇒\%m_{ancol}=49,46\%$
b/ 37,2 g hh có: 18,8 g phenol; 18,4 gam ancol
$C_6H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2$
$⇒n_{H_2}=\dfrac{1}{2}.(n_{C_6H_5OH}+n_{C_2H_5OH}=0,5.(0,2+0,4)=0,3\ mol$
⇒ 12,4g hh pư với Na tạo: $0,3.\dfrac{12,4}{37,3}=0,1\ mol\ H_2$
$V_{H_2}=2,24\ lít$
