Дано
m(ppa BaCL2)=15 g
W(BaCL2)=5%
H2SO4
-----------------------
n(BaSO4)-?
m(BaCL2)=15*5% / 100% = 0.75 g
0.75 X
BaCL2+H2SO4-->2HCL+BaSO4
208 233
M(BaCL2)=208g/mol , M(BaSO4)=233 g/mol
0.75 / 208 = X / 233
X = 0.84 g
n(BaSO4)= m/M=0.84 / 233 = 0.0036 mol
ответ 0.0036 моль