$a) NaHCO_3 + HCl → NaCl + H_2O + CO_2$
$\hspace{1cm}x\hspace{2,1cm}x\hspace{4,6cm}x$
$NaHSO_3 + HCl → NaCl + H_2O + SO_2$
$\hspace{0,5cm}y\hspace{2cm}y\hspace{4,8cm}y$
$n_{HCl}=0,35.1=0,35 (mol)$
$⇒x+y=0,35\quad (1)$
$m_{NaHCO_3, NaHSO_3}=33,4 (g)$
$⇒84x+104y=33,4\quad(2)$
$\text{Từ (1), (2)}⇒\left\{{{x+y=0,35}\atop{84x+104y=33,4}}\right.$
$⇔\left\{{{x=0,15}\atop{y=0,2}}\right.$
$⇒m_{NaHCO_3}=0,15.84=12,6 (g)$
$⇒m_{NaHSO_3}=33,4-12,6=20,8 (g)$
$b) n_{CO_2,SO_2}=x+y=0,35$
$⇒\%V_{CO_2}=\frac{0,15}{0,35}.100=42,86\%$
$⇒\%V_{SO_2}=100-42,86=57,14\%$