Giải thích các bước giải:

\(\begin{array}{l}
Zn + 2AgN{O_3} \to Zn{(N{O_3})_2} + 2Ag(1)\\
Zn + Cu{(N{O_3})_2} \to Zn{(N{O_3})_2} + Cu(2)\\
{n_{Zn}} = 0,25mol\\
{n_{AgN{O_3}}} = {n_{Cu{{(N{O_3})}_2}}} = 0,2mol\\
 \to {n_{Zn}}(1) = \dfrac{1}{2}{n_{AgN{O_3}}} = 0,1mol\\
 \to {n_{Ag}} = {n_{AgN{O_3}}} = 0,2mol\\
 \to {n_{Zn}}(2) = 0,25 - 0,1 = 0,15mol\\
 \to {n_{Zn}}(2) < {n_{Cu{{(N{O_3})}_2}}} \to {n_{Cu{{(N{O_3})}_2}}}dư\\
 \to {n_{Cu}} = {n_{Zn}}(2) = 0,15mol\\
a)\\
{m_{Ag}} = 0,2 \times 108 = 21,6g\\
{m_{Cu}} = 0,15 \times 64 = 9,6g\\
b)\\
Zn{(N{O_3})_2} + 2NaOH \to 2NaN{O_3} + Zn{(OH)_2}\\
{n_{NaOH}} = 0,8mol\\
{n_{Zn{{(N{O_3})}_2}}} = \dfrac{1}{2}{n_{AgN{O_3}}} + {n_{Zn}}(2) = 0,25mol\\
 \to {n_{Zn{{(N{O_3})}_2}}} < {n_{NaOH}} \to {n_{NaOH}}dư\\
 \to {n_{Zn{{(OH)}_2}}} = {n_{Zn{{(N{O_3})}_2}}} = 0,25mol\\
 \to {m_{Zn{{(OH)}_2}}} = 24,75g\\
\end{array}\)