nCaO = \(\dfrac{5,6}{56}\)= 0,1 mol
CaO + H2O -> Ca(OH)2
nCa(OH)2 = nCaO = 0,1 mol
nCO2 = \(\dfrac{2,8}{22,4}\)= 0,125 mol
Xét T = \(\dfrac{n_{CO2}}{n_{Ca\left(OH\right)2}}\)= \(\dfrac{0,125}{0,1}\)= 1,25 => sp tạo 2 muối
CO2 + Ca(OH)2 -> CaCO3\(\downarrow\) + H2O (1)
..x...........x.................x
2CO2 + Ca(OH)2 -> Ca(HCO3)2 (2)
..y............0,5y.............0,5y
Từ (1) và (2) ta có hệ \(\left\{{}\begin{matrix}x+y=0,125\\x+0,5y=0,1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x=0,075\\y=0,05\end{matrix}\right.\)
=> mkết tủa = 0,075.100 = 7,5 g