Đáp án:
20,18% 79,82%
Giải thích các bước giải:
\(\begin{array}{l}
RCHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to RCOON{H_4} + N{H_4}N{O_3} + 2Ag\\
nAg = \dfrac{{38,88}}{{108}} = 0,36\,mol\\
nRCHO = \dfrac{{0,36}}{2} = 0,18\,mol\\
MA = \dfrac{{9,81}}{{0,18}} = 54,5g/mol \Rightarrow MR = 25,5\\
\Rightarrow CTPT:C{H_3}CHO(a\,mol),{C_2}{H_5}CHO(b\,mol)\\
44a + 58b = 9,81\\
a + b = 0,18\\
\Rightarrow a = 0,045;b = 0,135\\
\% mC{H_3}CHO = \dfrac{{0,045 \times 44}}{{9,81}} \times 100\% = 20,18\% \\
\% m{C_2}{H_5}CHO = 100 - 20,18 = 79,82\%
\end{array}\)