Дано
m(ppa H2SO4)=245 g
W(H2SO4)=20%
BaCL2
------------------------
m(BaSO4)-?
m(H2SO4)=m(ppaH2SO4)*W(H2SO4)/100% = 245*20%/100% = 49 g
H2SO4+BaCl2-->2HCL+BaSO4↓
M(H2SO4)=98 g/mol
n(H2SO4)=m/M=49/98 = 0.5 mol
n(H2SO4)=n(BaSO4) = 0.5 mol
M(BaSO4)=233g/mol
m(BaSO4) = n(BaSO4)*M(BaSO4)= 0.5*233= 116.5 g
ответ 116.5 г