Đáp án:
\({m_{C{H_3}COOH}} = 12{\text{ gam}}\)
\({m_{C{H_3}COO{C_2}{H_5}}} = 17,6{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{{C_2}{H_5}OH}} = \frac{{9,2}}{{46}} = 0,2{\text{ mol = }}{{\text{n}}_{C{H_3}COOH}} = {n_{C{H_3}COO{C_2}{H_5}}}\)
\( \to {m_{C{H_3}COOH}} = 0,2.60 = 12{\text{ gam}}\)
\({m_{C{H_3}COO{C_2}{H_5}}} = 0,2.88 = 17,6{\text{ gam}}\)