Đáp án:
27,71% và 72,29%
Giải thích các bước giải:
\(\begin{array}{l}
CTTQ:{C_{\overline n }}{H_{2\overline n + 1}}OH\\
2{C_{\overline n }}{H_{2\overline n + 1}}OH + 2Na \to 2{C_{\overline n }}{H_{2\overline n + 1}}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
\Rightarrow {n_{{C_{\overline n }}{H_{2\overline n + 1}}OH}} = 2{n_{{H_2}}} = 0,3\,mol\\
{M_{{C_{\overline n }}{H_{2\overline n + 1}}OH}} = \dfrac{{16,6}}{{0,3}} = 55,33g/mol\\
\Rightarrow 14\overline n + 18 = 55,33 \Leftrightarrow \overline n = 2,67\\
\Rightarrow hh:{C_2}{H_5}OH(a\,mol),{C_3}{H_7}OH(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
46a + 60b = 16,6
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
\% {m_{{C_2}{H_5}OH}} = \dfrac{{0,1 \times 46}}{{16,6}} \times 100\% = 27,71\% \\
\% {m_{{C_3}{H_7}OH}} = 100 - 27,71 = 72,29\%
\end{array}\)
