Đáp án:

\(\begin{array}{l}
a)\\
{m_{AgCl}} = 14,35g\\
b)\\
C{\% _{Cu{{(N{O_3})}_2}}} = 5,56\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CuC{l_2} + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2AgCl\\
{n_{CuC{l_2}}} = \dfrac{m}{M} = \dfrac{{6,75}}{{135}} = 0,05mol\\
{n_{AgCl}} = 2{n_{CuC{l_2}}} = 0,1mol\\
{m_{AgCl}} = n \times M = 0,1 \times 143,5 = 14,35g\\
b)\\
{m_{ddCuC{l_2}}} = \dfrac{{6,75 \times 100}}{{13,5}} = 50g\\
{n_{AgN{O_3}}} = {n_{AgCl}} = 0,1mol\\
{m_{AgN{O_3}}} = n \times M = 0,1 \times 170 = 17g\\
{m_{ddAgN{O_3}}} = \dfrac{{17 \times 100}}{{15}} = 133,3g\\
{m_{ddspu}} = 50 + 133,3 - 14,35 = 168,95g\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{CuC{l_2}}} = 0,05mol\\
{m_{Cu{{(N{O_3})}_2}}} = n \times M = 0,05 \times 188 = 9,4g\\
C{\% _{Cu{{(N{O_3})}_2}}} = \dfrac{{9,4}}{{168,95}} \times 100\%  = 5,56\% 
\end{array}\)