\(n_{K_2O}=\dfrac{28,2}{94}=0,3\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{230.9,8\%}{98}=0,23\left(mol\right)\)
PTHH: K2O + H2SO4 --> K2SO4 + H2O
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,23}{1}\) => H2SO4 hết, K2O dư và pư với nước
PTHH: K2O + H2SO4 --> K2SO4 + H2O
0,23<----0,23------>0,23
K2O + H2O --> 2KOH
0,07----------->0,14
mdd sau pư = 28,2 + 230 = 258,2 (g)
\(\left\{{}\begin{matrix}m_{K_2SO_4}=0,23.174=40,02\left(g\right)\\m_{KOH}=0,14.56=7,84\left(g\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{40,02}{258,2}.100\%=15,5\%\\C\%_{KOH}=\dfrac{7,84}{258,2}.100\%=3,04\%\end{matrix}\right.\)