Đáp án:
Câu 3:
a) ancol etylic: 4,6 g; axit axetic: 3 g
b) 2,64 g
Giải thích các bước giải:
Câu 2:
1. $2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}$
2. $C{H_3}COOH + NaHC{O_3} \to C{H_3}COONa + C{O_2} + {H_2}O$
3. $2C{H_3}COOH + CaO \to {(C{H_3}COO)_2}Ca + {H_2}O$
Câu 3:
a) $C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O$
$\begin{gathered}
\Rightarrow {n_{C{H_3}COOH}} = {n_{NaOH}} = 0,05mol \hfill \\
\Rightarrow {m_{C{H_3}COOH}} = 0,05.60 = 3g \hfill \\
\end{gathered} $
$ \Rightarrow {m_{{C_2}{H_5}OH}} = 7,6 - 3 = 4,6g$
b) $C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O$
Do ${n_{{C_2}{H_5}OH}} = 0,1mol > {n_{C{H_3}COOH}} = 0,05mol$ ⇒ Tính theo $_{C{H_3}COOH}$
$\begin{gathered}
H = 60\% \Rightarrow {n_{C{H_3}COOH(pu)}} = 0,05.H = 0,03mol \hfill \\
\Rightarrow {n_{este}} = {n_{C{H_3}COOH(pu)}} = 0,03mol \hfill \\
\end{gathered} $
$ \Rightarrow {m_{este}} = 0,03.88 = 2,64g$
