Gọi số mol K, Ba, K2O là a, b, c
=> 39a + 137b + 94c = 15,61 (1)
PTHH: 2K + 2H2O --> 2KOH + H2
_____a---------------->a----->0,5a
Ba + 2H2O --> Ba(OH)2 + H2
b---------------->b--------->b
K2O + H2O --> 2KOH
c---------------->2c
=> 0,5a + b = \(\dfrac{2,016}{22,4}=0,09\left(mol\right)\) (2)
PTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O
_______b--------->b--------->b
2KOH + CO2 --> K2CO3 + H2O
(a+2c)->(0,5a+c)->(0,5a+c)
K2CO3 + CO2 + H2O --> 2KHCO3
(0,5a+c)->(0,5a+c)
=> \(\left\{{}\begin{matrix}b=\dfrac{V}{22,4}\\b+\left(a+2c\right)=\dfrac{V+4,48}{22,4}\end{matrix}\right.\)
=> a + 2c = 0,2 (3)
=> \(\left\{{}\begin{matrix}a=0,08\\b=0,05\\c=0,06\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{KOH}=0,2\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,05\left(mol\right)\end{matrix}\right.\)