$V_{C_2H_5OH} = 100.\dfrac{92}{100} = 92(ml)$
$d_{C_2H_5OH} = 0,8(g/ml) \Rightarrow m_{C_2H_5OH} = 0,8.92 = 73,6(gam)$
$n_{C_2H_5OH} = 1,6(mol)$
Mỗi phần, $n_{C_2H_5OH} = 1,6 : 2 = 0,8(mol)$
Phần 1 :
$C_6H_5OH + 3Br_2 \to C_6H_2Br_3OH + 3HBr$
$n_{C_6H_5OH} = n_{kết\ tủa} = \dfrac{39,72}{331} = 0,12(mol)$
Phần 2 :
$2C_6H_5OH + 2Na \to 2C_6H_5ONa + H_2$
$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$
$n_{H_2} = \dfrac{1}{2}n_{phenol} + \dfrac{1}{2}n_{C_2H_5OH} = 0,46(mol)$
$V_{H_2} = 0,46.22,4 = 10,304(lít)$