Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{{C_2}{H_5}OH}} = 31,5\% \\
\% {m_{{C_4}{H_8}{O_2}}} = 63,5\% \\
b){m_{{C_2}{H_5}ONa}} = 6,8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_5}OH + Na \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,05mol\\
a)\\
{n_{{C_2}{H_5}OH}} = 2{n_{{H_2}}} = 0,1mol\\
\to {m_{{C_2}{H_5}OH}} = 4,6g\\
\to \% {m_{{C_2}{H_5}OH}} = \dfrac{{4,6}}{{14,6}} \times 100\% = 31,5\% \\
\to \% {m_{{C_4}{H_8}{O_2}}} = 100\% - 31,5\% = 63,5\% \\
b)\\
{n_{{C_2}{H_5}ONa}} = 2{n_{{H_2}}} = 0,1mol\\
\to {m_{{C_2}{H_5}ONa}} = 6,8g
\end{array}\)
