Đáp án:
1, \({C_2}{H_5}OH\)
2, \({C_3}{H_7}OH\)
3, \(C{H_3}OH,{C_2}{H_5}OH\)
Giải thích các bước giải:
1,
\(\begin{array}{l}
ROH + Na \to RONa + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,125mol\\
\to {n_{ROH}} = 2{n_{{H_2}}} = 0,25mol\\
\to {M_{ROH}} = \dfrac{{11,5}}{{0,25}} = 46 \to {C_2}{H_5}OH
\end{array}\)
2,
\(\begin{array}{l}
{n_{C{O_2}}} = 0,09mol\\
{n_{{H_2}O}} = 0,12mol\\
{n_{{H_2}O}} > {n_{C{O_2}}}
\end{array}\)
Suy ra ancol no, đơn chức
\(\begin{array}{l}
\to {n_{ancol}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,03mol\\
\to C = \dfrac{{{n_{C{O_2}}}}}{{{n_{ancol}}}} = 3\\
\to {C_3}{H_7}OH
\end{array}\)
3,
\(\begin{array}{l}
ROH + Na \to RONa + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,2mol\\
\to {n_{ROH}} = 2{n_{{H_2}}} = 0,4mol\\
\to {M_{ROH}} = 42,5\\
\to C{H_3}OH,{C_2}{H_5}OH
\end{array}\)
