Đáp án:
\(\% {m_{C{H_3}CHO}} = 33,13\% ;\% {m_{{C_2}{H_5}COOH}} = 66,87\% \)
m=33,2 gam
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_3}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O\xrightarrow{{}}C{H_3}COON{H_4} + 2N{H_4}N{O_3} + 2Ag\)
\( \to {n_{Ag}} = \frac{{54}}{{108}} = 0,5{\text{ mol}} \to {{\text{n}}_{C{H_3}CHO}} = \frac{1}{2}{n_{Ag}} = 0,25{\text{ mol}}\)
Cho X tác dụng với natri cacbonat
\(2{C_2}{H_5}COOH + N{a_2}C{O_3}\xrightarrow{{}}2{C_2}{H_5}COONa + C{O_2} + {H_2}O\)
\({n_{C{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{n}}_{{C_2}{H_5}COOH}} = 2{n_{C{O_2}}} = 0,3{\text{ mol}}\)
\( \to {m_{C{H_3}CHO}} = 0,25.44 = 11{\text{ gam;}}{{\text{m}}_{{C_2}{H_5}COOH}} = 0,3.74 = 22,2{\text{ gam}} \to {\text{m = 11 + 22}}{\text{,2 = 33}}{\text{,2 gam}}\)
\( \to \% {m_{C{H_3}CHO}} = \frac{{11}}{{33,2}} = 33,13\% \to \% {m_{{C_2}{H_5}COOH}} = 66,87\% \)
Ta có: \(\frac{{17,4}}{{33,2}} = 0,524\)
Vậy 17,4 gam X chứa 0,524.0,25=0,131 mol andehit
\(C{H_3}CHO + B{r_2} + {H_2}O\xrightarrow{{}}C{H_3}COOH + 2HBr\)
\( \to {n_{B{r_2}}} = {n_{C{H_3}CHO}} = 0,131{\text{ mol}} \to {{\text{m}}_{B{r_2}}} = 0,131.160 = 20,96{\text{ gam}}\)
