BaCl2 + H2SO4 => BaSO4↓ + 2HCl
mBaCl2=5,2.400/100=20,8g
=> nBaCl2=20,8/208=0,1mol
mdd H2SO4=100.1,14=114g
=> mH2SO4=22,8g
=> nH2SO4=22,8/98=57/245mol
Vì 0,1 < 57/245 => BaCl2 hết , H2SO4 dư = 57/245 - 0,1 =13/98mol
=> mH2SO4(dư)=13/98 . 98 = 13g
=> mdd(spư)=400+114=514g
=> C%H2SO4=13/514 . 100%=2,529%
CmH2SO4=13/98:0,1=65/49M
=> C%HCl=7,3/514 . 100%=1,420%
CmHCl=0,2/0,1=2M
CHÚC BẠN HỌC TỐT :)))