$Na_2CO_3 + HCl → 2NaCl + H_2O + CO_2$
$a\hspace{7,5cm}a$
$NaHCO_3 + HCl → NaCl + H_2O + CO_2$
$b\hspace{7,5cm}b$
$m_{Na_2CO_3, NaHCO_3}=9,5 (g)$
$⇒106a+84b=9,5\quad (1)$
$n_{CO_2}=\frac{2,24}{22,4}=0,1 (mol)$
$⇒a+b=0,1\quad (2)$
Từ (1), (2)$⇒\left\{{{106a+84b=9,5}\atop{a+b=0,1}}\right.$
$⇔a=b=0,05$
$⇒m_{Na_2CO_3}=0,05.106=5,3 (g)$
$⇒m_{NaHCO_3}=9,5-5,3=4,2 (g)$