Em tham khảo nha :
\(\begin{array}{l}
a)\\
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,224}}{{22,4}} = 0,01mol\\
{n_{hh}} = 2{n_{{H_2}}} = 0,02mol\\
M = \dfrac{{1,06}}{{0,02}} = 53dvC\\
\Rightarrow 14n + 18 = 53\\
\Rightarrow n = \dfrac{{53 - 18}}{{14}} = 2,5\\
\Rightarrow CTPT:{C_2}{H_5}OH;{C_3}{H_7}OH\\
b)\\
hh:{C_2}{H_5}OH(a\,mol),{C_3}{H_7}OH(b\,mol)\\
\left\{ \begin{array}{l}
46a + 60b = 1,06\\
a + b = 0,02
\end{array} \right.\\
\Rightarrow a = 0,01;b = 0,01\\
{m_{{C_2}{H_5}OH}} = 0,01 \times 46 = 0,46g\\
\% {C_2}{H_5}OH = \dfrac{{0,46}}{{1,06}} \times 100\% = 43,4\% \\
\% {C_3}{H_7}OH = 100 - 43,4 = 56,6\%
\end{array}\)
