\(n_{CO_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
\(n_{KOH}=0,8.0,05=0,04\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,02.0,8=0,016\left(mol\right)\)
PTHH: Ca(OH)2 + CO2 --> CaCO3\(\downarrow\) + H2O
______0,016---->0,016---->0,016___________(mol)
2KOH + CO2 --> K2CO3 + H2O
0,04--->0,02----->0,02_____________________(mol)
K2CO3 + CO2 + H2O --> 2KHCO3
0,004<---0,004----------->0,008_______________(mol)
mCaCO3 = 0,016.100 = 1,6 (g)
A chứa K2CO3: 0,016 mol và KHCO3: 0,008 mol
\(\left\{{}\begin{matrix}C_{M\left(K_2CO_3\right)}=\dfrac{0,016}{0,8}=0,02M\\C_{M\left(KHCO_3\right)}=\dfrac{0,008}{0,8}=0,01M\end{matrix}\right.\)