Đáp án:
3,69% 96,31%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2{C_3}{H_7}OH + 2Na \to 2{C_3}{H_7}ONa + {H_2}\\
b)\\
n{H_2} = \dfrac{{2,352}}{{22,4}} = 0,105\,mol\\
hh:{C_2}{H_5}OH(a\,mol),{C_3}{H_7}OH(b\,mol)\\
46a + 60b = 12,46\\
0,5a + 0,5b = 0,105\\
\Rightarrow a = 0,01;b = 0,2\\
\% m{C_2}{H_5}OH = \dfrac{{0,46}}{{12,46}} \times 100\% = 3,69\% \\
\% m{C_3}{H_7}OH = 100 - 3,69 = 96,31\%
\end{array}\)
