K2O + H2O \(\rightarrow\)2KOH (1)
a;
nK2O=\(\dfrac{9,4}{94}=0,1\left(mol\right)\)
Theo PTHH 1 ta có:
2nK2O=nKOH=0,2(mol)
CM dd KOH=\(\dfrac{0,2}{0,2}=1M\)
b;
2KOH + H2SO4 \(\rightarrow\)K2SO4 + 2H2O (2)
Theo PTHH 2 ta có:
\(\dfrac{1}{2}\)nKOH=nH2SO4=0,1(mol)
mH2SO4=0,1.98=9,8(g)
mdd H2SO4=9,8:\(\dfrac{20}{100}\)=49(g)
Vdd H2SO4=49:1,15=42,6(ml)