Em tham khảo nha:

\(\begin{array}{l}
T{H_1}: \text{ Chỉ có 1 muối tạo kết tủa }\\
 \Rightarrow X:Flo,Y:Clo \Rightarrow hh:NaF,NaCl\\
AgN{O_3} + NaCl \to AgCl + NaN{O_3}\\
{n_{AgCl}} = \dfrac{{57,34}}{{143,5}} \approx 0,4\,mol\\
{n_{NaCl}} = {n_{AgCl}} = 0,4\,mol\\
\% {m_{NaCl}} = \dfrac{{0,4 \times 58,5}}{{31,84}} \times 100\%  = 73,49\% \\
\% {m_{NaF}} = 100 - 73,49 = 26,51\% \\
T{H_2}: \text{ Cả 2 muối đều tạo kết tủa }\\
NaM + AgN{O_3} \to AgM + NaN{O_3}\\
m = (108 - 23) \times {n_M} \Leftrightarrow 85{n_M} = 57,34 - 31,84\\
 \Rightarrow {n_M} = 0,3\,mol\\
{M_M} = \dfrac{{31,84}}{{0,3}} - 23 = 83,13g/mol\\
 \Rightarrow X:Brom(Br),Y:Iot(I)\\
NaBr + AgN{O_3} \to AgBr + NaN{O_3}\\
NaI + AgN{O_3} \to AgI + NaN{O_3}\\
hh:NaBr(a\,mol),NaI(b\,mol)\\
\left\{ \begin{array}{l}
103a + 150b = 31,84\\
188a + 235b = 57,34
\end{array} \right.\\
 \Rightarrow a = 0,28;b = 0,02\\
\% {m_{NaBr}} = \dfrac{{0,28 \times 103}}{{31,84}} \times 100\%  = 90,58\% \\
\% {m_{NaI}} = 100 - 90,58 = 9,42\% 
\end{array}\)