PTHH :
K2O + H2O -> 2KOH
0,2.......................0,4 (mol)
nK2O = \(\dfrac{18,8}{94}=0,2\left(mol\right)\)
mKOH = 0,4.56=22,4(g)
mKOH 7,63% = 0,0763m (g)
=>\(\Sigma m_{KOH-trong-dung-dịch}=22,4+0,0763m\)
Ta có hệ :
\(\dfrac{22,4+0,0763m}{22,4+m}=21\%\)
=> m\(\approx132,36\left(g\right)\)