Giải thích các bước giải:
3,
Gọi 2 ancol có CT là: ROH
\(\begin{array}{l}
ROH + Na \to RONa + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{ROH}} = 2{n_{{H_2}}} = 0,3mol\\
\to {M_{ROH}} = 55,33 \to R = 38,3\\
\to {C_2}{H_5},{C_3}{H_7}\\
\to {C_2}{H_5}OH,{C_3}{H_7}OH
\end{array}\)
Vậy 2 ancol đó là \({C_2}{H_5}OH,{C_3}{H_7}OH\)
4,
\(\begin{array}{l}
{n_{C{O_2}}} = 0,26mol\\
{n_{{H_2}O}} = 0,36mol\\
{n_{{H_2}O}} > {n_{C{O_2}}}\\
{n_{ancol}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,1mol\\
\to C = \dfrac{{{n_{C{O_2}}}}}{{{n_{ancol}}}} = 2,6
\end{array}\)
\( \to {C_2}{H_5}OH,{C_3}{H_7}OH\)
Gọi a và b là số mol của \({C_2}{H_5}OH,{C_3}{H_7}OH\)
\(\begin{array}{l}
{C_2}{H_6}O \to 2C{O_2} + 3{H_2}O\\
{C_3}{H_8}O \to 3C{O_2} + 4{H_2}O\\
\left\{ \begin{array}{l}
2a + 3b = 0,26\\
3a + 4b = 0,36
\end{array} \right.\\
\to a = 0,04 \to b = 0,06\\
\to {m_{{C_2}{H_5}OH}} = 1,84g \to {m_{{C_3}{H_7}OH}} = 3,6g
\end{array}\)
\(\begin{array}{l}
{C_2}{H_6}O + CuO \to C{H_3}CHO + {H_2}O + Cu\\
{C_3}{H_8}O + CuO \to C{H_3}C{H_2}CHO + {H_2}O + Cu\\
C{H_3}CHO \to 2Ag\\
C{H_3}C{H_2}CHO \to 2Ag
\end{array}\)
