Em tham khảo nha :
\(\begin{array}{l}
1)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{n_{NaOH}} = 0,3 \times 1 = 0,3mol\\
{n_{C{H_3}COOH}} = {n_{NaOH}} = 0,3mol\\
{n_{{C_2}{H_5}OH}} = 2 \times (0,4 - \dfrac{{0,3}}{2}) = 0,5mol\\
m = {m_{C{H_3}COOH}} + {m_{{C_2}{H_5}OH}} = 0,3 \times 60 + 0,5 \times 46 = 41g\\
2)\\
4AgN{O_3} + 3C{H_3}CHO + 5N{H_3} \to 4Ag + 3N{H_4}N{O_3} + 3C{H_3}CN{H_4}\\
{n_{C{H_3}CHO}} = \dfrac{{4,4}}{{44}} = 0,1mol\\
{n_{Ag}} = \dfrac{4}{3}{n_{C{H_3}CHO}} = \dfrac{2}{{15}}mol\\
{m_{Ag}} = \dfrac{2}{{15}} \times 108 = 14,4g
\end{array}\)
