Giải thích các bước giải:
\(\begin{array}{l}
a){C_2}{H_5}OH + C{H_3}{\rm{COO}}H \to C{H_3}{\rm{COO}}{C_2}{H_5} + {H_2}O\\
{n_{_{C{H_3}{\rm{COO}}{C_2}{H_5}}}} = 0,3mol\\
\to {n_{{C_2}{H_5}OH}} = {n_{C{H_3}{\rm{COO}}H}} = 0,3mol\\
\to {m_{{C_2}{H_5}OH}} = 0,3 \times 46 = 13,8g\\
\to {m_{C{H_3}{\rm{COO}}H}} = 0,3 \times 60 = 18g\\
{n_{{C_2}{H_5}OH(lt)}} = 1mol\\
{n_{C{H_3}{\rm{COO}}H(lt)}} = 0,5mol\\
\to {n_{{C_2}{H_5}OH(dư)}} = 1 - 0,3 = 0,7mol\\
\to {n_{C{H_3}{\rm{COO}}H(dư)}} = 0,5 - 0,3 = 0,2mol\\
b)\\
{m_{{C_2}{H_5}OH(dư)}} = 0,7 \times 46 = 32,2g\\
{m_{C{H_3}{\rm{COO}}H(dư)}} = 0,2 \times 60 = 12g\\
c)\\
H\% = \dfrac{{32,2}}{{46}} \times 100\% = 70\%
\end{array}\)