Đáp án:
\( {m_{C{H_2} = C(C{H_3})COOH}} = 107,5{\text{ gam}}\)
\( {m_{C{H_3}OH}} = 40{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_2} = C(C{H_3})COOH + C{H_3}OH\xrightarrow{{{H_2}S{O_4};{t^o}}}C{H_2} = C(C{H_3})COOC{H_3} + {H_2}O\)
Ta có:
\({n_{C{H_2} = C(C{H_3})COOC{H_3}}} = \dfrac{{100}}{{100}} = 1{\text{ mol}}\)
\( \to {n_{C{H_2} = C(C{H_3})COOH{\text{ lt}}}} = {n_{C{H_3}OH{\text{ lt}}}} = {n_{este}} = 1{\text{ mol}}\)
Tuy nhiên hiệu suất chỉ là $80\%.$
\( \to {n_{C{H_2} = C(C{H_3})COOH}} = {n_{C{H_3}OH}} = \dfrac{1}{{80\% }} = 1,25{\text{ mol}}\)
\(\to {m_{C{H_2} = C(C{H_3})COOH}} = 1,25.86 = 107,5{\text{ gam}}\)
\( \to {m_{C{H_3}OH}} = 1,25.32 = 40{\text{ gam}}\)
