Đáp án:
4) ${V_{kk}} = 168(l)$; ${m_{C{O_2}}} = 44g;{m_{{H_2}O}} = 27g$
5) $\% {m_{C{H_3}COOH}} = 83,91\% ;\% {m_{ancol}} = 16.09\% $
6) ${C_3}{H_7}OH$
Giải thích các bước giải:
4) ${C_2}{H_6}O + 3{O_2} \to 2C{O_2} + 3{H_2}O$
${m_{{C_2}{H_6}O}} = D.V = 28,75.0,8 = 23g \to {n_{{C_2}{H_6}O}} = 0,5mol$
Theo PTHH:
$\begin{gathered}
{n_{C{O_2}}} = 2{n_{{C_2}{H_6}O}} = 1mol \to {m_{C{O_2}}} = 44g \hfill \\
{n_{{H_2}O}} = 3{n_{{C_2}{H_6}O}} = 1,5mol \Rightarrow {m_{{H_2}O}} = 27g \hfill \\
\end{gathered} $
$\begin{gathered}
{n_{{O_2}}} = 3{n_{{C_2}{H_6}O}} = 1,5mol \hfill \\
\Rightarrow {V_{kk}} = 5{V_{{O_2}}} = 5.1,5.22,5 = 168(l) \hfill \\
\end{gathered} $
5) $C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O$
${n_{C{H_3}COOH}} = {n_{NaOH}} = 0,1.2 = 0,2mol$
$\begin{gathered}
\Rightarrow {m_{C{H_3}COOH}} = 0,2.60 = 12g \hfill \\
\Rightarrow \% {m_{C{H_3}COOH}} = \dfrac{{12}}{{14,3}}.100\% = 83,91\% \hfill \\
\Rightarrow \% {m_{ancol}} = 100 - 83,91 = 16.09\% \hfill \\
\end{gathered} $
6)
$C{H_3}COOH + ROH \to C{H_3}COOR + {H_2}O$
$\begin{gathered}
{n_{C{H_3}COOH(bd)}} = \dfrac{6}{{60}} = 0,1mol \hfill \\
H = 70\% \Rightarrow {n_{C{H_3}COOH(pu)}} = 0,07mol \hfill \\
\Rightarrow {n_{este}} = {n_{C{H_3}COOH(pu)}} = 0,07mol \hfill \\
\end{gathered} $
$\begin{gathered}
{M_{este}} = \dfrac{{7,14}}{{0,07}} = 102 \hfill \\
\Rightarrow R + 59 = 102 \Rightarrow R = 43 \hfill \\
\end{gathered} $
⇒ R là ${C_3}{H_7} - $
CTHH của ancol là ${C_3}{H_7}OH$
