Đáp án:

$\rm a)\\V_{C_2H_5OH}=25×69\%=17,25\ (mL)\\m_{C_2H_5OH}=17,25×0,8=13,8\ (g)\\b)\\m_{H_2O}=V_{H_2O}=25-17,25=7,75\ (mL)\\n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\ (mol)\\n_{H_2O}=\dfrac{7,75}{18}=0,43\ (mol)$

Phản ứng xảy ra:

$\rm C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2\\H_2O+Na\to NaOH+\dfrac{1}{2}H_2$

Theo phản ứng:

$\rm n_{H_2}=\dfrac{1}{2}×(n_{C_2H_5OH}+n_{H_2O})=0,365\ (mol)$

$\rm V_{H_2}=0,365×22,4=8,176\ (L)$