a,PTHH : \(K_2O+H_2O\rightarrow2KOH\)
\(n_{K_2O}=\frac{m}{M}=\frac{9,4}{39.2+16}=0,1\left(mol\right)\)
- Theo PTHH : \(n_{KOH}=2n_{K_2O}=2.0,1=0,2\left(mol\right)\)
-> \(C_M=\frac{n}{V}=\frac{0,2}{0,5}=0,4\left(M\right)\)
b, - Để phản ứng tạo ra cả hai muối .
<=> \(1< T< 2\)
<=> \(\left\{{}\begin{matrix}\frac{n_{KOH}}{n_{SO_2}}>1\\\frac{n_{KOH}}{n_{SO_2}}< 2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\frac{0,2}{n_{SO_2}}>1\\\frac{0,2}{n_{SO_2}}< 2\end{matrix}\right.\)
<=> \(0,1< n_{SO_2}< 0,2\left(mol\right)\)
Vậy ....