Дано
m(H2SO4)=20 g
m(ppa BaCL2)=150 g
W=10%
-----------------------------
m(BaSO4)-?
m(в-ва BaCL2)=150*10%/100%=15 g
20 15 X
H2SO4+BaCL2-->2HCL+BaSO4
98 208 233
M(H2SO4)=98 g/mol M(BaCL2)=208 g/mol M(BaSO4)=233 g/mol
n(H2SO4)=m/M=20/98=0.2 mol
n(BaCL2)=15/208=0.07 mol
n(H2SO4)>n(BaCL2)
15/208 = X/233
X=16.8 g
ответ 16.8 г