\(\text{Ba(OH)2+H2SO4->BaSO4+2H2O}\)
Ta có :
\(\text{nBaSO4=nBa(OH)2=17,1/171=0,1(mol) }\)
\(\Rightarrow\text{mdd spu=17,1+200-0,1x233=193,8(g) }\)
nBa(OH)2=\(\frac{17,1}{171}\)=0,1(mol)
Pt: Ba(OH)2+H2SO4→BaSO4+2H2O
0,1mol →0,1mol→0,1mol →0,2mol
mBaSO4=0,1.233=23,3g
mH2O=0,2.18=3,6g
mH2SO4 dư =200-0,1.98=190,2g
Σ mdd spu=190,2+3,6+23,3=217,1g