2KOH + H2SO4 -> K2SO4 + 2H2O
mKOH=200.2%=4g => nKOH=1/14 mol
nH2SO4=0,2.1,5=0,3 mol => mH2SO4=29,4g
Vì (1/14)/2<0,3/1 => Sau pứ H2SO4 dư
nK2SO4=nKOH=1/14 mol => mK2SO4=12,43g
mdd sau pứ=mddKOH + mdd H2SO4=200+29,4=229,4g
C%K2SO4=12,43/229,4.100%=5,42%
nH2SO4 pứ=1/28 mol
mH2SO4 dư=(0,3-1/28).98=25,9g
C%H2SO4=25,9/229,4.100%=11,29%