+ Phần 2:
nNaOH = 0,1.1 = 0,1 (mol)
PTHH: C6H5OH + NaOH --> C6H5ONa + H2O
0,1<------0,1
+ Phần 1:
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2C6H5OH + 2Na --> 2C6H5ONa + H2
0,1---------------------------->0,05
2C6H5CH2OH + 2Na --> 2C6H5CH2ONa + H2
0,1<-------------------------------------0,05
=> hh ban đầu chứa \(\left\{{}\begin{matrix}C_6H_5OH:0,2\left(mol\right)\\C_6H_5CH_2OH:0,2\left(mol\right)\end{matrix}\right.\)
=> m = 0,2.94 + 0,2.108 = 40,4 (g)
=> D