Đáp án:
\(\begin{array}{l}
{m_{{C_6}{H_5}OH}} = 4,7g\\
{m_{{C_2}{H_5}OH}} = 11,5g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}(1)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}(2)\\
{C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{NaOH}} = 0,1 \times 0,5 = 0,05mol\\
{n_{{C_6}{H_5}OH}} = {n_{NaOH}} = 0,05mol\\
{n_{{H_2}(2)}} = \dfrac{{{n_{{C_6}{H_5}OH}}}}{2} = 0,025mol\\
{n_{{H_2}(1)}} = 0,15 - 0,025 = 0,125ml\\
{n_{{C_2}{H_5}OH}} = 2{n_{{H_2}(1)}} = 0,25mol\\
{m_{{C_6}{H_5}OH}} = 0,05 \times 94 = 4,7g\\
{m_{{C_2}{H_5}OH}} = 0,25 \times 46 = 11,5g
\end{array}\)
