`n_{CaO}=\frac{2,8}{56}=0,05(mol)`
`n_{HCl}=0,2(mol)`
`a)` `CaO+2HCl\to CaCl_2+H_2O`
Do `\frac{0,05}{1}=\frac{0,2}{2}`
`=>` axit dư
`=> n_{HCl\ pứ}=2n{CaO}=0,1(mol)`
`m_{HCl}=0,1.36,5=3,65g`
`n_{HCl\ dư}=0,2-0,1=0,1(mol)`
`b)` `n_{CaCl_2}=n_{CaO}=0,05(mol)`
`=> m_{\text{muối}}=m_{CaCl_2}=0,05.111=5,55g`
`c)``m_{dd\ HCl}=D.V=1,25.200=250g`
`m_{ddspu}=2,8+250=252,8g`
`C%_{HCl\ dư}=\frac{0,1.36,5.100%}{252,8}\approx 1,44%`
`C%_{CaCl_2}=\frac{5,55.100%}{252,8}\approx 2,2%`