$n_{SO_2}=\frac{3,36}{22,4}=0,15 mol$
$n_{NaOH}= 0,1.1,7=0,17 mol$
$n_{NaOH} : n_{SO_2}=1,13$
$\Rightarrow $ Tạo 2 muối: $Na_2SO_3$ (a mol), $NaHSO_3$ (b mol)
Bảo toàn Na: $2a+b=0,17$ (1)
Bảo toàn S: $a+b=0,15$ (2)
(1)(2) $\Rightarrow a=0,02; b=0,13$
$m_{\text{muối}}= 0,02.126+0,13.104=16,04g$