Đáp án:

$\rm a)$

Phương trình phản ứng:

$\rm CH_3CH_2OH+Na\to CH_3CH_2OH+\dfrac{1}{2}H_2\\C_6H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\\C_6H_5OH+NaOH\to C_6H_5ONa+H_2O$

$\rm b)\\n_{H_2}=\dfrac{7,437}{24,79}=0,3\ (mol)\\n_{NaOH}=0,35×1=0,35\ (mol)$

Theo phương trình:

$\rm n_{C_6H_5OH}=n_{NaOH}=0,35\ (mol)$

$\rm n_{C_2H_5OH}=2n_{H_2}-n_{C_6H_5OH}=0,25\ (mol)$

Phần trăm khối lượng:

$\rm \%m_{C_6H_5OH}=\dfrac{0,35×94}{0,35×94+0,25×46}×100\%=74,1\%\\\%m_{C_2H_5OH}=(100-74,1)\%=25,9\%$