Đáp án:
\(\begin{array}{l}
a)\\
CTPT:C{H_3}OH,{C_2}{H_5}OH\\
b)\\
\% {m_{C{H_3}OH}} = 41,03\% \\
\% {m_{{C_2}{H_5}OH}} = 58,97\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_X} = 2{n_{{H_2}}} = 0,4\,mol\\
{M_X} = \dfrac{{15,6}}{{0,4}} = 39g/mol \Rightarrow 14n + 18 = 39\\
\Rightarrow n = 1,5 \Rightarrow CTPT:C{H_3}OH,{C_2}{H_5}OH\\
b)\\
hh\,X:C{H_3}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,4\\
32a + 46b = 15,6
\end{array} \right.\\
\Rightarrow a = b = 0,2\,mol\\
\% {m_{C{H_3}OH}} = \dfrac{{0,2 \times 32}}{{15,6}} \times 100\% = 41,03\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 41,03 = 58,97\%
\end{array}\)
