mKOH(bđ)=10%.200=20(g)
Gọi a là số mol K2O cần thêm
=>\(m_{K_2O}\)=94a
K2O+H2O->2KOH
a.......................2a.....(mol)
Theo PTHH:mKOH(pt)=2a.56=112a
=>mKOH=20+112a(g)
Ta có:mdd(sau)=94a+200(g)
Theo gt:C%KOH=12%
=>\(\dfrac{20+112a}{200+94a}\).100%=12%
=>2000+11200a=2400+1128a
=>10072a=400=>a=0,04(mol)
=>\(m_{K_2O}\)=94a=94.0,04=3,76(g)