Đáp án:

\(16.B\)

\(17.D\)

\(18.D\)

\(19.C\)

\(20.C\)

Giải thích các bước giải:

\(16.B\)

\(n_{KOH}^{} = 0,15.1 = 0,15\;{\text{mol}}\)

\({\text{Este don chuc mach ho nen}}\)

\({\text{n}}_{KOH}^{} = n_{Este}^{} = 0,15\)

\(M_X^{} = \frac{{12,9}}{{0,15}} = 86g/mol\)

\( \to C_4^{}H_6^{}O_2^{}\)

\({\text{khi phan ung xong duoc 1 muoi va 1 andehit}}\)

\( \to {\text{X la CH}}_3^{}COOC_2^{}H_3^{}\)

\(17.D\)

\(n_C^{} = \frac{9}{{12}};n_O^{} = \frac{8}{{16}}\)

\(C:O = \frac{9}{{12}}:\frac{8}{{16}} = 3:2\)

\( \to C_3^{}H_X^{}O_2^{}\)

\(\frac{{Muoi}}{{Este}} = \frac{{41}}{{37}}\)

\(Neu\;{\text{muoi la CH}}_3^{}COONa \to este = 74\)

\( \to CH_3^{}COOCH_3^{}\)

\(18.D\)

\(n_{HCl}^{} = 0,2\;{\text{mol}}\)

\(n_{NaOH_{PU}^{}}^{} = 0,5 - 0,2 = 0,3\;{\text{mol}}\)

\({\text{3 NaOH tao ra 1 ancol}}\)

\( \to {\text{M}}_{Muoi}^{} = \frac{{24,6}}{{0,3}} = 82g/mol\)

\(Muoi\;{\text{la CH}}_3^{}COONa\)

\(M_A^{} = 218g/mol\)

\(Ruou\;{\text{B la C}}_3^{}H_5^{}(OH)_3^{}\)

\(A\;{\text{la C}}_3^{}H_5^{}(OOCCH_3^{})_3^{}\)

\(19.C\)

\(M = 44.2 = 88g/mol\)

\(44,6g{\text{ chat ran}}\;{\text{co NaOH}}_{du}^{}{\text{ = 0}}{\text{,8 - 0}}{\text{,4 = 0}}{\text{,4 mol va Muoi cua este}}\)

\({\text{co the la HCOO va CH}}_3^{}COO\)

\(88x + 88y = 35,2\)

\(68x + 82y + 40.0,4 = 44,6\)

\( \to x = 0,3;y = 0,1\)

\({\text{hoac CH}}_3^{}COO{\text{ va C}}_2^{}H_5^{}COO\)

\(88x + 88y = 35,2\)

\(96x + 82y + 40.0,4 = 44,6\)

\( \to x =  - 0,3;y = 0,7(Loai)\)

\(20.C\)

\(n_{KOH}^{} = 0,075.2 = 0,15\;{\text{mol}}\)

\({\text{M}}_{TrungBinh}^{} = \frac{{10,4}}{{0,15}} = 69,3g/mol\)

\({\text{ chac chan co }}HCOOCH_3^{}\)

\(2\;{\text{muoi khan dong dang ke tiep va 1 ancol}}\)

\(HCOOCH_3^{}\& CH_3^{}COOCH_3^{}\)