$n_{NaOH}=0,07.0,5=0,035 mol$
Gọi a, b là mol $Na_2SO_3$, $NaHSO_3$.
$\Rightarrow 126a+104b=3,435$ (1)
$2NaOH+SO_2\to Na_2SO_3+2H_2O$
$NaOH+SO_2\to NaHSO_3$
$\Rightarrow 2a+b=0,035$ (2)
(1)(2)$\Rightarrow a=0,0025; b=0,03$
$\Rightarrow n_{SO_2}=0,0025+0,03=0,0325 mol$
Gọi x, y là mol Ag, Cu.
$\Rightarrow 108x+64y=4,32$ (*)
$2Ag+2H_2SO_4\to Ag_2SO_4+SO_2+2H_2O$
$Cu+2H_2SO_4\to CuSO_4+SO_2+2H_2O$
$\Rightarrow 0,5x+y=0,0325$ (**)
(*)(**)$\Rightarrow x=0,029; y=0,007$
$\%Ag=\dfrac{0,029.108.100}{4,32}=72,5\%$
$\%Cu=27,5\%$