`mH_2SO_4= 29,4.20%=5,88`(g)
`=> nH_2SO_4=0,06` (mol)
`mBaCl_2=100.5,2%=5,2` (g)
`=> nBaCl_2 =0,025` (mol)
a, `BaCl_2 + H_2SO_4 -> BaSO_4↓ + 2HCl`
`nBaCl_2 < nH_2SO_4 (0,025<0,06 )`
`=> BaCl_2` hết `; H_2SO_4` dư
`=> nBaSO_4 = nBaCl_2 = 0,025` (mol)
`=> mBaSO_4=5,825` (g)
b,
`mdd = m`dd `BaCl_2 + m`dd `H2SO_4 - mBaSO_4`
`= 29,4+100-5,825 `
`=123,575`
`nHCl = 2.nBaCl_2= 0,05` (mol)
`=> mHCl = 1,825` (g)
`nH_2SO_4` pứ `= nBaCl2 = 0,025` (mol)
`mH_2SO_4` dư `= (0,06-0,025).98=3,43` (g)
`% H_2SO_4=(3,43)/(123,575).100%=2,776%`
`%HCl = (1,825)/(123,575).100%=1,477%`
