Đáp án:
79,968l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2{H_2}O + 2Na \to 2NaOH + {H_2}\\
b)\\
{V_{{C_2}{H_5}OH}} = 250 \times \dfrac{{75}}{{100}} = 187,5ml\\
{m_{{C_2}{H_5}OH}} = 187,5 \times 0,9 = 168,75g\\
{m_{{H_2}O}} = \dfrac{{250 - 187,5}}{1} = 62,5g\\
{n_{{C_2}{H_5}OH}} = \dfrac{{168,75}}{{46}} = \dfrac{{675}}{{184}}\,mol\\
{n_{{H_2}O}} = \dfrac{{62,5}}{{18}} = \dfrac{{125}}{{36}}\,mol\\
{n_{{H_2}}} = \dfrac{{675}}{{184}} \times \dfrac{1}{2} + \dfrac{{125}}{{36}} \times \dfrac{1}{2} = 3,57mol\\
{V_{{H_2}}} = 3,57 \times 22,4 = 79,968l
\end{array}\)
