\(K_2O+H_2O-->2KOH\)
\(KOH+HCl->KCl+H_2O\)
n(K2O0=9,4/94=0,1 mol=> nKOH=0,2 mol
=> mHCl=0,2.36,5=7,3 g
mddHCl=7,3.100/30=24,33333gam
V=m/D=24,3333/1,5=16,2222ml
\(K_2O(0,1)+H_2O--->2KOH(0,2)\)\((1)\) \(nH_2O=0,1(mol)\) Theo PTHH (1) \(nKOH=0,2\left(mol\right)\) \(KOH(0,2)+HCl(0,2)--->KCl+H_2O\)\((2)\) Theo PTHH (2) \(nHCl=0,2(mol)\) \(\Rightarrow m_{HCl}=7,3\left(g\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{7,3.100}{30}=\dfrac{73}{3}\left(g\right)\) \(\Rightarrow V_{ddHCl}=16,22\left(l\right)\)