\(\begin{array}{l}
1)\\
500ml=0,5l\to n_{AgNO_3}=0,3.0,5=0,15(mol)\\
Cu+2AgNO_3\to Cu(NO_3)_2+2Ag\\
Tang\,giam\,khoi\,luong:\\
\to n_{Cu}=\frac{22,56-13,44}{108.2-64}=0,06(mol)\\
Theo\,PT:\,n_{Cu(NO_3)_2}=n_{Cu}=0,06(mol)\\
n_{AgNO_3(du)}=0,15-0,06.2=0,03(mol)\\
\to \begin{cases} C_{M\,Cu(NO_3)_2}=\frac{0,06}{0,5}=0,12M\\ C_{M\,AgNO_3(du)}=\frac{0,03}{0,5}=0,06M \end{cases}\\
2)\\
Dat\,hoa\,tri\,R\,la\,n(n\in\mathbb{N^*})\\
2R+nCu(NO_3)_2\to 2R(NO_3)_n+nCu\\
R+nAgNO_3\to R(NO_3)_n+nAg\\
n_{Ag}=n_{AgNO_3(du)}=0,03(mol)\\
n_{Cu}=n_{Cu(NO_3)_2}=0,06(mol)\\
\sum n_R=\frac{2n_{Cu}}{n}+\frac{n_{Ag}}{n}=\frac{0,15}{n}(mol)\\
\Delta m=17,205-15=2,205(g)\\
\to m_{Cu}+m_{Ag}-m_R=2,205\\
\to 0,03.108+0,06.64-\frac{0,15R}{n}=2,205\\
\to \frac{0,15R}{n}=4,875\\
\to R=32,5n\\
\to \begin{cases} n=2\\ R=65 \end{cases}\\
\to R:Zn/kem
\end{array}\)