\(\text{mH2SO4=200.9,8%=19,6g}\)
\(\Rightarrow\text{nH2SO4=0,2mol. nNaOH=0,15mol.}\)
\(\text{PTHH: 2NaOH+H2SO4=Na2SO4+2H2O}\)
\(\text{2KOH+H2SO4=K2SO4+2H2O. }\)
nNaOH=0,15mol => nH2SO4 (1) = 0,075mol
=> nH2SO4 (2) =0,2-0,075=0,125mol
=> nKOH=0,25mol
=> mKOH=0,25.56=14g
\(\text{Vậy C% KOH= 14.100:100=14% }\)