Đáp án:

$\rm a)$

Phản ứng xảy ra:

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

$\rm b)$

$\rm V_{C_2H_5OH}=10×96\%=9,6\ (mL)$

$\rm m_{C_2H_5OH}=9,6×0,8=7,68\ (g)$

$\rm c)\\n_{C_2H_5OH}=\dfrac{7,68}{46}=\dfrac{96}{575}\ (mol)\\n_{H_2O}=\dfrac{10-9,6}{18}=\dfrac{1}{45}\ (mol)$

Theo phương trình:

$\rm n_{H_2}=\dfrac{1}{2}(n_{C_2H_5OH}+n_{H_2O})=\dfrac{979}{10350}\ (mol)$

$\rm V_{H_2}=\dfrac{979}{10350}×22,4=2,1188\ (L)$