n Ba(OH)2=$\frac{17,1}{171}$ =0,1 mol
Ba(OH)2+H2SO4→BaSO4↓+2H2O
0,1→ 0,1 0,1 mol
a.m ct H2SO4=0,1.98=9,8 g
C% H2SO4=$\frac{9,8}{200}$ .100=4,9%
b.mdd sau=m Ba(OH)2+m H2SO4
=17,1+200=217,1 g
c.C% BaSO4=$\frac{233.0,1}{217,1}$ .100≈10,732 %
...............................chúc bạn học tốt..........................