`a)`
Xuất hiện kết tủa trắng
`CaSO_4+BaCl_2\to BaSO_4+CaCl_2`
`b)`
`n_{CaSO_4}={1,11}/{136}=0,0082`mol
`n_{BaCl_2}={2,33}/{208}=0,011`mol
`=>BaCl_2` dư
`n_{BaSO_4}=n_{CaSO_4}=0,0082`mol
`=>m_↓=233.0,0082=1,9106g`
`c)`
`V_{dd}=100ml=0,1l`
`n_{CaCl_2}=n_{CaSO_4}=0,0082`mol
`=> C_{M_{CaCl_2}}={0,0082}/{0,1}=0,082M`
`n_{BaCl_2}=0,011-0,0082=0,0028`mol
`=> C_{M_{BaCl_2}}={0,0028}/{0,1}=0,028M`