Đặt $n_{CO_2}=x(mol);\,n_{SO_2}=y(mol)$
$CO_2+Ca(OH)_2\to CaCO_3+H_2O\\\,\,\,\,x\quad\quad\quad x\quad\quad\quad\quad\quad x\quad\quad\quad\quad\quad(mol)$
$SO_2+Ca(OH)_2\to CaSO_3+H_2O\\\,\,\,\,y\quad\quad\quad y\quad\quad\quad\quad\quad y\quad\quad\quad\quad\quad(mol)$
$\sum n_{hh}=\dfrac{19,6}{22,4}=0,875(mol)\\↔x+y=0,875(mol)$
$\sum m_{muoi}=100x+120y=90(g)$
$→$ Ta có hệ phương trình: $\begin{cases}x+y=0,875\\100x+120y=90\end{cases}↔\begin{cases}x=0,75\\y=0,125\end{cases}$
hay $\begin{cases}n_{CO_2}=0,75(mol)\\n_{SO_2}=0,125(mol)\end{cases}$
$→\begin{cases}V_{CO_2}=16,8(l)\\V_{SO_2}=2,8(l)\end{cases}$
$\sum n_{Ca(OH)_2}=x+y=0,875(mol)\\→C_{M_{Ca(OH)_2}}=\dfrac{0,875}{0,2}=4,375(M)$
Vậy $V_{CO_2}=16,8l;\,V_{SO_2}=2,8l;\,C_{M_{Ca(OH)_2}}=4,375M$
