Em tham khảo nha :
\(\begin{array}{l}
a)\\
NaCl + AgN{O_3} \to NaN{O_3} + AgCl\\
KCl + AgN{O_3} \to KN{O_3} + AgCl\\
{n_{AgCl}} = \dfrac{{2,87}}{{143,5}} = 0,02mol\\
hh:NaCl(a\,mol),KCl(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,02\\
58,5a + 74,5b = 1,33
\end{array} \right.\\
\Rightarrow a = 0,01;b = 0,01\\
{m_{NaCl}} = 0,01 \times 58,5 = 0,585g\\
{m_{NaC{l_{bd}}}} = 0,585 \times 10 = 5,85g\\
{m_{KC{l_{bd}}}} = 13,3 - 5,85 = 7,45g\\
b)\\
C{\% _{NaCl}} = \dfrac{{5,85}}{{500}} \times 100\% = 1,17\% \\
C{\% _{KCl}} = \dfrac{{7,45}}{{500}} \times 100\% = 1,49\%
\end{array}\)
