Đáp án:

\(\begin{array}{l}
a)\\
{m_{NaCl}} = 5,85g\\
{m_{KCl}} = 7,45g\\
b)\\
{C_\% }NaCl = 1,17\% \\
{C_\% }KCl = 1,49\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
NaCl + AgN{O_3} \to AgCl + NaN{O_3}\\
KCl + AgN{O_3} \to AgCl + KN{O_3}\\
{n_{AgCl}} = \dfrac{{2,87}}{{143,5}} = 0,02\,mol\\
hh:NaCl(a\,mol),KCl(b\,mol)\\
\left\{ \begin{array}{l}
58,5a + 74,5b = 13,3\\
0,1a + 0,1b = 0,02
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
{m_{NaCl}} = 0,1 \times 58,5 = 5,85g\\
{m_{KCl}} = 13,3 - 5,85 = 7,45g\\
b)\\
{C_\% }NaCl = \dfrac{{5,85}}{{500}} \times 100\%  = 1,17\% \\
{C_\% }KCl = \dfrac{{7,45}}{{500}} \times 100\%  = 1,49\% 
\end{array}\)