Đáp án:
Câu 30: 114,5 g
Câu 31: 10,9%
Giải thích các bước giải:
Câu 30:
${C_6}{H_5}OH + 3HN{O_3}\xrightarrow{{{H_2}S{O_{4d}}}}{C_6}{H_2}{(N{O_2})_3}OH + 3{H_2}O$
${n_{{C_6}{H_5}OH}} = \dfrac{{47}}{{94}} = 0,5mol$
${n_{HN{O_3}}} = \dfrac{{200.68\% }}{{63}} = 2,16mol$
Ta có: $\dfrac{{{n_{{C_6}{H_5}OH}}}}{1} < \dfrac{{{n_{HN{O_3}}}}}{3}$ ⇒ ${_{HN{O_3}}}$ dư; ${_{{C_6}{H_5}OH}}$ hết
$\begin{gathered}
\Rightarrow {n_{axit{\text{ picric}}}} = {n_{phenol}} = 0,5mol \hfill \\
\Rightarrow {m_{axit{\text{ picric}}}} = 0,5.229 = 114,5g \hfill \\
\end{gathered} $
Câu 31:
${n_{NaOH}} = 0,2.0,5 = 0,1mol$
${C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O$
$\begin{gathered}
\Rightarrow {n_{{C_6}{H_5}OH}} = {n_{NaOH}} = 0,1mol \hfill \\
\Rightarrow {m_{{C_6}{H_5}OH}} = 0,1.94 = 9,4g \hfill \\
\end{gathered} $
$ \Rightarrow {m_{{C_2}{H_5}OH}} = 47 - 9,4 = 37,6g$
$\begin{gathered}
\Rightarrow {n_{{C_2}{H_5}OH}} = \dfrac{{37,6}}{{46}} = 0,817mol \hfill \\
\Rightarrow \% {n_{{C_6}{H_5}OH}} = \dfrac{{0,1}}{{0,817 + 0,1}}.100\% = 10,9\% \hfill \\
\end{gathered} $
