\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Ba\left(OH\right)_2}=1,5.0,1=0,15\left(mol\right)\\ 1< \dfrac{n_{CO_2}}{n_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,15}< 2\)
=> Tạo 2 muối
\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\\a.........a..........a\left(mol\right)\\ Ba\left(OH\right)_2+2CO_2\rightarrow Ba\left(HCO_3\right)_2\\ b...........2b.............b\left(mol\right)\\ \left\{{}\begin{matrix}a+2b=0,2\\a+b=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ m_{\downarrow}=197a=19,7\left(g\right)\)
$n_{Ba(OH)_2} = 0,15(mol) ; n_{CO_2} = 0,2(mol)$
Ba(OH)2 + CO2 → BaCO3 + H2O
0,15.............0,15........0,15.............................(mol)
BaCO3 + CO2 + H2O → Ba(HCO3)2
0,05...........0,05...........................................(mol)
$m_{kết\ tủa} = (0,15 - 0,05).197 = 19,7(gam)$
