Đáp án:
\(m=16,94 gam\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_3}COO{C_2}{H_5} + NaOH\xrightarrow{{}}C{H_3}COONa + {C_2}{H_5}OH\)
\({H_2}NC{H_2}COOH + NaOH\xrightarrow{{}}{H_2}NC{H_2}COONa + {H_2}O\)
Ta có:
\({n_{C{H_3}COO{C_2}{H_5}}} = \frac{{4,4}}{{88}} = 0,05{\text{ mol;}}{{\text{n}}_{{H_2}NC{H_2}COOH}} = \frac{9}{{75}} = 0,12{\text{ mol}}\)
\({n_{NaOH}} = 0,1.2 = 0,2{\text{ mol > }}{{\text{n}}_{C{H_3}COO{C_2}{H_5}}} + {n_{{H_2}NC{H_2}COOH}}\)
Vậy \(NaOH\) dư
\({n_{{C_2}{H_5}OH}} = {n_{C{H_3}COO{C_2}{H_5}}} = 0,05{\text{ mol}}\)
\({n_{{H_2}O}} = {n_{{H_2}NC{H_2}COOH}} = 0,12{\text{ mol}}\)
BTKL:
\({m_{C{H_3}COO{C_2}{H_5}}} + {m_{{H_2}NC{H_2}COOH}} + {m_{NaOH}} = m + {m_{{C_2}{H_5}OH}} + {m_{{H_2}O}}\)
\( \to 4,4 + 9 + 0,2.40 = m + 0,05.46 + 0,12.18\)
\( \to m=16,94 gam\)
