Em tham khảo nha :
\(\begin{array}{l}
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}\\
{n_{{C_n}{H_{2n + 1}}OH}} = \dfrac{6}{{14n + 18}}mol\\
{n_{{C_n}{H_{2n + 1}}ONa}} = \dfrac{{8,2}}{{14n + 40}}mol\\
{n_{{C_n}{H_{2n + 1}}OH}} = {n_{{C_n}{H_{2n + 1}}ONa}}\\
\frac{6}{{14n + 18}} = \dfrac{{8,2}}{{14n + 40}}\\
\Rightarrow n = 3\\
\Rightarrow CTPT:{C_3}{H_7}OH\\
b)\\
C{H_3} - C{H_2} - C{H_2} - OH:propan - 1 - ol\\
C{H_3} - CH(OH) - C{H_3}:propan - 2 - ol
\end{array}\)
