Đáp án:
\( {V_{{H_2}}} = 3,584{\text{ lít}}\)
\( {m_{dd\;{\text{C}}{{\text{H}}_3}COOH}} = 384{\text{ gam}}\)
\( {m_{C_2H_5OH}} = 12,512{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + Zn\xrightarrow{{}}{(C{H_3}COO)_2}Zn + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{10,4}}{{65}} = 0,16{\text{ mol}}\)
\({n_{{H_2}}} = {n_{Zn}} = 0,16{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0.16.22.4 = 3,584{\text{ lít}}\)
\( \to {n_{C{H_3}COOH}} = 2{n_{Zn}} = 0,32{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,32.60 = 19,2{\text{ gam}}\)
\( \to {m_{dd\;{\text{C}}{{\text{H}}_3}COOH}} = \frac{{19,2}}{{5\% }} = 384{\text{ gam}}\)
Thực hiện este hóa:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
\( \to {n_{C_2H_5OH_{lt}}}= {n_{C{H_3}COOH}} = 0,32{\text{ mol}}\)
\( \to {n_{C_2H_5OH}} = 0,32.85\% = 0,272{\text{ mol}}\)
\(\to {m_{C_2H_5OH}} = 0,272.46 = 12,512{\text{ gam}}\)
