Em tham khảo nha:
\(\begin{array}{l}
5)\\
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
{C_6}{H_{12}}{O_6} + AgN{O_3} + N{H_3} \to 2Ag + 2N{H_4}N{O_3} + {C_6}{H_{12}}{O_7}\\
6)\\
a)\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
A:C{H_3}COOH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
60a + 46b = 10,6\\
0,5a + 0,5b = 0,1\\
\Rightarrow a = b = 0,1\,mol\\
\% mC{H_3}COOH = \dfrac{{0,1 \times 60}}{{10,6}} \times 100\% = 56,6\% \\
\% m{C_2}{H_5}OH = 100 - 56,6 = 43,4\% \\
b)\\
n{C_2}{H_5}ONa = n{C_2}{H_5}OH = 0,1\,mol\\
nC{H_3}COONa = nC{H_3}COOH = 0,1\,mol\\
m{C_2}{H_5}ONa = 0,1 \times 68 = 6,8g\\
mC{H_3}COONa = 0,1 \times 82 = 8,2g
\end{array}\)
