nCO2 = \(\dfrac{3,36}{22,4}\)= 0,15 mol
nBa(OH)2 = 1.0,125 = 0,125 mol
Xét T = \(\dfrac{n_{CO2}}{n_{Ba\left(OH\right)2}}\)= 1,2
Xảy ra pt:
CO2 + Ba(OH)2 -> BaCO3\(\downarrow\)+ H2O (1)
..x...........x................x
2CO2 + Ba(OH)2 -> Ba(HCO3)2 (2)
..2y..........y.................y
Từ (1) và (2) ta có hệ: \(\left\{{}\begin{matrix}x+2y=0,15\\x+y=0,125\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=0,1\\y=0,025\end{matrix}\right.\)
CM BaCO3 = \(\dfrac{0,1}{0,125}\)= 0,8M
CM Ba(HCO3)2 = \(\dfrac{0,025}{0,125}\)= 0,2M