Gọi a, b là mol phenol, etanol mỗi phần.
- P1:
$n_{H_2}=0,3 mol$
$2ROH+2Na\to 2RONa+H_2$
$\Rightarrow 0,5a+0,5b=0,3$ (1)
- P2:
$n_{KOH}=\frac{25.1,4.40\%}{56}=0,25 mol$
$C_6H_5OH+KOH\to C_6H_5OK+H_2O$
$\Rightarrow a=0,25$ (2)
(1)(2) $\Rightarrow b=0,35 mol$
$m=2.(0,25.94+0,35.46)=79,2g$