\(\begin{array}{l}
2)\\
nS{O_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
nKOH = 0,5 \times 0,45 = 0,225\,mol\\
\dfrac{{nKOH}}{{nS{O_2}}} = \dfrac{{0,225}}{{0,15}} = 1,5\\
 =  > {\rm{dd}}:{K_2}S{O_3},KHS{O_3}\\
b)\\
2KOH + S{O_2} \to {K_2}S{O_3} + {H_2}O(1)\\
{K_2}S{O_3} + S{O_2} + {H_2}O \to 2KHS{O_3}(2)\\
nS{O_2}(1) = n{K_2}S{O_3}(1) = 0,1125\,mol\\
nS{O_2}(2) = n{K_2}S{O_3}(2) = 0,15 - 0,1125 = 0,0375\,mol\\
nKHS{O_3} = 0,0375 \times 2 = 0,075\,mol\\
n{K_2}S{O_3} = 0,1125 - 0,0375 = 0,075\,mol\\
m = mKHS{O_3} + m{K_2}S{O_3} = 20,85g\\
3)\\
NaM + AgN{O_3} \to AgM + NaN{O_3}\\
nNaM = nAgM\\
 =  > \dfrac{{6,03}}{{23 + MM}} = \dfrac{{8,61}}{{108 + MM}}\\
 =  > MM = 175,66
\end{array}\)

=> chỉ có 1 chất cho kết tủa

=>NaCl và NaF

\(\begin{array}{l}
nNaCl = nAgCl = \dfrac{{8,61}}{{143,5}} = 0,06\,mol\\
\% mNaCl = \dfrac{{0,06 \times 58,5}}{{6,03}} \times 100\%  = 58,21\% \\
\% mNaF = 100 - 58,21 = 41,79\% 
\end{array}\)