\(\begin{array}{l}
CO_2+Ca(OH)_2\to CaCO_3+H_2O\\
Theo\,PT:\,n_{CaCO_3}=n_{CO_2}=\frac{20}{100}=0,2(mol)\\
\overline{M_X}=20,4.2=40,8(g/mol)\\
\to \frac{n_{CO(du)}}{n_{CO_2}}=\frac{44-40,8}{40,8-28}=\frac{1}{4}\\
\to n_{CO(du)}=\frac{1}{4}n_{CO_2}=0,05(mol)\\
Dat\,n_{FeO}=x(mol);n_{Fe_2O_3}=y(mol)\\
Bao\,toan\,C:\,n_{CO(pu)}=n_{CO_2}=0,2(mol)\\
Bao\,toan\,KL:\\
m_A=m_B+m_{CO_2}-m_{CO(pu)}\\
\to m_A=12+0,2.44-0,2.28=15,2(g)\\
\to 72x+160y=15,2(1)\\
Ma\,x+y=0,15(2)\\
(1)(2)\to x=0,1;y=0,05\\ 
\to \%m_{FeO}=\frac{0,1.72}{15,2}.100\%\approx 47,37\%\\
\to \%m_{Fe_2O_3}=100-47,37=52,63\%\\
\sum n_{CO}=0,2+0,05=0,25(mol)\\
\to V=0,25.22,4=5,6(l)
\end{array}\)