Lời giải:Đốt cháy E \(\xrightarrow{{}}6,75 + 0,3975.32 = 44{n_{C{O_2}}} + 0,275.18\xrightarrow{{}}{n_{C{O_2}}} = 0,33\)
\(\xrightarrow{{BTKL}}n_O^{trong\,E} = 0,14\xrightarrow{{}}{n_E} = {n_{COO}} = 0,07\)
Ta có: \({n_{NaOH}} = 0,07\xrightarrow{{}}{n_E} = 0,07\xrightarrow{{}}{n_{ancol}} = 0,07\xrightarrow{{chay}}\left\{ \begin{gathered}
C{O_2}:a \hfill \\
{H_2}O:a + 0,07 \hfill \\
\end{gathered} \right.\)
\(\xrightarrow{{BTNT.O}}a = 0,12\xrightarrow{{}}n_C^{axit} = 0,21\xrightarrow{{}}{C_3}\) và \(\left\{ \begin{gathered}
C{H_3}OH:0,02 \hfill \\
{C_2}{H_5}OH:0,05 \hfill \\
\end{gathered} \right.\)
\(\xrightarrow{{}}n_H^{trong\,RCOO} = 0,24\xrightarrow{{}}\overline H = 3,43\xrightarrow{{}}\left\{ \begin{gathered}
{C_2}{H_5}COO:0,015 \hfill \\
{C_2}{H_3}COO:0,055 \hfill \\
\end{gathered} \right.\xrightarrow{{}}3:11\)
