Đáp án:
$\begin{gathered}
{m_{C{H_3}COOH(du)}} = 100,8g;{m_{{C_2}{H_5}OH(du)}} = 119,278g \hfill \\
{m_{este}} = 116,16g \hfill \\
\end{gathered} $
Giải thích các bước giải:
${n_{C{H_3}COOH}} = \dfrac{{180}}{{60}} = 3mol;{n_{{C_2}{H_5}OH}} = \dfrac{{180}}{{46}} = 3,913mol$
$C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O$
$\begin{gathered}
\Rightarrow {n_{C{H_3}COOH(pu)}} = {n_{{C_2}{H_5}OH(pu)}} = {n_{este}} = 3.\dfrac{{44}}{{100}} = 1,32mol \hfill \\
\Rightarrow {n_{C{H_3}COOH(du)}} = 3 - 1,32 = 1,68mol \hfill \\
{n_{{C_2}{H_5}OH(du)}} = 3,913 - 1,32 = 2,593mol \hfill \\
\Rightarrow {m_{C{H_3}COOH(du)}} = 1,68.60 = 100,8g \hfill \\
{m_{{C_2}{H_5}OH(du)}} = 2,593.46 = 119,278g \hfill \\
{m_{este}} = 1,32.88 = 116,16g \hfill \\
\end{gathered} $
