$\rm a) Dr =\dfrac{V_{C_2H_5OH}}{V_{dd}}.100=\dfrac{30}{90}.100=33,3^0$ 

$\rm b)$ 

$\rm PTHH : 2C_2H_5OH +2Na \xrightarrow{} 2C_2H_5ONa +H_2$ 

                   $\rm 2Na +2H_2O \xrightarrow{} 2NaOH +H_2$

$m_{C_2H_5OH}=0,8.30 =24(g)$ 

`->` $n_{C_2H_5OH}=\dfrac{24}{46}=\dfrac{12}{23}(mol)$ 

$m_{H_2O}=1.90=90(g)$ 

`->`$n_{H_2O}=\dfrac{90}{18}=5(mol)$ 

theo pthh 

$n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+ \dfrac{1}{2}n_{H_2O}=\dfrac{6}{23} +2,5 =\dfrac{127}{46}(mol)$ 

`->`$V_{H_2}=\dfrac{127}{46}.22,4 =61,84(l)$