Câu 4:
a,
$n_{CO_2}= 0,3 mol$
$n_{H_2O}= 0,425 mol$
CTTQ 2 ancol là $C_nH_{2n+2}O$
Ta có: $n_{ancol}=\frac{0,3}{n}=\frac{0,425}{n+1}$
$\Leftrightarrow n= 2,4$
Vậy 2 ancol là $C_2H_5OH$ (a mol), $C_3H_7OH$ (b mol)
b,
Bảo toàn C: $2a+3b=0,3$ (1)
Bảo toàn H: $3a+4b=0,425$ (2)
(1)(2) $\Rightarrow a=0,075; b=0,05$
$\%m_{C_2H_5OH}=\frac{0,075.46.100}{0,075.46+0,05.60}= 53,49\%$
$\%m_{C_3H_7OH}= 46,51\%$
Câu 5:
a,
$2C_2H_5OH+2Na \to 2C_2H_5ONa+ H_2$
$2C_6H_5OH+2Na \to 2C_6H_5ONa+ H_2$
$C_6H_5OH+ NaOH \to C_6H_5ONa+ H_2O$
b,
$n_{H_2}= 0,1 mol$
$n_{NaOH}= 0,1.0,5=0,05 mol$
Gọi a, b là mol etanol, phenol
$\Rightarrow 0,5a+0,5b=0,1; b=0,05$
$\Leftrightarrow a=0,15; b=0,05$
$m=0,15.46+0,05.94= 11,6g$
$\%m_{C_2H_5OH}=\frac{0,15.46.100}{11,6}= 59,48\%$
$\%m_{C_6H_5OH}= 40,52\%$
