giúp mình 5 câu này vs ạ ..

Đáp án:

\(11.C\)

\(12.A\)

\(13.B\)

\(14.C\)

\(15.B\)

Giải thích các bước giải:

\(11.C\)

\(X{\text{ la este don chuc}}\)

\({\text{n}}_X^{} = n_{KOH}^{} = 0,5.0,1 = 0,05\;{\text{mol}}\)

\({\text{M}}_X^{} = \frac{{5,8}}{{0,05}} = 116g/mol\)

\(12m + 2m + 1 + 73 = 116\)

\(m = 3\)

\( \to C_3^{}H_7^{}COOC_2^{}H_5^{}\)

\(12.A\)

\(n_{NaOH}^{} = \frac{{12\% .100}}{{40}} = 0,3\;{\text{mol}}\)

\(M_{Ancol}^{} = \frac{{9,2}}{1} = 92g/mol \to C_3^{}H_5^{}{\text{(OH)}}_3^{}\)

\( \to n_{Muoi}^{} = 0,3\;{\text{mol}}\)

\({\text{M}}_{Muoi}^{} = \frac{{20,4}}{{0,3}} = 68g/mol\)

\( \to Axit:HCOOH\)

\(13.B\)

\(M_{C_2^{}H_5^{}OH}^{} = 62\% M_X^{}\)

\( \to X = 74g/mol\)

\( \to HCOOC_2^{}H_5^{}\)

\(14.C\)

\(n_{O_2^{}}^{} = n_X^{} = \frac{{3,2}}{{32}} = 0,1\;{\text{mol}}\)

\({\text{M}}_X^{} = \frac{{7,4}}{{0,1}} = 74g/mol\)

\(\frac{{RCOONa}}{{74}} = \frac{{41}}{{37}} \to R = 15 \to CH_3^{}COOCH_3^{}\)

\(15.B\)

\(d_E^{} = 29.4 = 116g/mol \to C_6^{}H_{12}^{}O_2^{}\)

\(\frac{{RCOONa}}{{116}} = \frac{{24}}{{29}} \to R = 29\)

\( \to C_2^{}H_5^{}COOC_3^{}H_7^{}\)