BaCL2+H2SO4-->2HCL+BaSO4 208 233 M(BaCL2)=208 g/mol n(BaCL2)= m/M= 52/208=0.25 mol M(H2SO4)=98 g/mol n(H2SO4)=m/M=49 / 98 = 0.5 mol n(. BaCl2)<n(H2SO4) M(BaSO4)=233 g/mol 52 / 208 = X/ 233 X= 52*233/208 = 58.25g ответ 58.25 г.
BaCL2+H2SO4-->2HCL+BaSO4 208 233 M(BaCL2)=208 g/mol n(BaCL2)= m/M= 52/208=0.25 mol M(H2SO4)=98 g/mol n(H2SO4)=m/M=49 / 98 = 0.5 mol n(. BaCl2)<n(H2SO4) M(BaSO4)=233 g/mol 52 / 208 = X/ 233 X= 52*233/208 = 58.25g ответ 58.25 г.