Đáp án + Giải thích các bước giải:

$V_{C_2H_5OH}=\frac{8}{0,8}=10(ml)$

$V_{H_2O}=10.1=10(ml)$

$\to Đ_r=\frac{10}{10+10}.100=50^o$

$n_{C_2H_5OH}=\frac{8}{46}=\frac{4}{23}(mol)$

$n_{H_2O}=\frac{10}{18}=\frac{5}{9}(mol)$

$2Na+2C_2H_5OH\to 2C_2H_5ONa+H_2$

$2Na+2H_2O\to 2NaOH+H_2$

Theo PT: $\sum n_{H_2}=\frac{1}{2}\left(n_{C_2H_5OH}+n_{H_2O}\right)=\frac{151}{414}(mol)$

$\to V_{H_2(đktc)}=\frac{151}{414}.22,4\approx 8,17(l)$

Em tham khảo nha :

\(\begin{array}{l}
{V_{{C_2}{H_5}OH}} = \dfrac{8}{{0,8}} = 10ml\\
{V_{{H_2}O}} = \dfrac{{10}}{1} = 10ml\\
{D_r} = \frac{{10}}{{10 + 10}} \times 100 = {50^0}\\
{n_{{C_2}{H_5}OH}} = \dfrac{8}{{46}} = \dfrac{4}{{23}}mol\\
{n_{{H_2}O}} = \dfrac{{10}}{{18}} = \dfrac{5}{9}mol\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{{n_{{C_2}{H_5}OH}}}}{2} + \dfrac{{{n_{{H_2}O}}}}{2} = 0,365mol\\
{V_{{H_2}}} = 0,365 \times 22,4 = 8,176l
\end{array}\)