Lời giải:

$\rm a)\\V_{C_2H_5OH}=25\cdot 57,5\%=14,375\ (mL)\\V_{H_2O}=25-14,375=10,625\ (mL)$

Số gam rượu nguyên chất:

$\rm m_{C_2H_5OH}=14,375\cdot 0,8=11,5\ (g)$

$\rm b)\\n_{C_2H_5OH}=\dfrac{11,5}{46}=0,25\ (mol)\\n_{H_2O}=\dfrac{10,625\cdot 1}{18}=0,59\ (mol)$

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:

$\rm n_{H_2}=\dfrac{1}{2}\cdot (n_{C_2H_5OH}+n_{H_2O})=0,42\ (mol)$

$\rm V_{H_2}=0,42\cdot 22,4=9,408\ (L)$

`a,` 

`V_{C_2H_5OH}=(57,5.25)/(100)=14,375(ml)`

`=>m_{C_2H_5OH}=14,375.0,8=11,5(g)`

`b,` `n_{C_2H_5OH}=(11,5)/(46)=0,25(mol)`

`V_{H_2O}=25-14,375=10,625(ml)`

`=>m_{H_2O}=10,625.1=10,625(g)`

`=>n_{H_2O}=(10,625)/(18)=(85)/(144)(mol)`

PTHH:
`2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2\uparrow`

`2H_2O+2Na\to 2NaOH+H_2\uparrow`

`=>n_{H_2}= 1/2 n_{C_2H_5OH}+ 1/2 n_{H_2O}`

            `=1/2 . 0,25+ 1/2 . (85)/(144)≈0,42(mol)`

`=>V_{H_2}=0,42.22,4=9,408(l)`