\(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\)
PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => CaO hết, HCl dư
Theo PTHH: \(n_{CaCl_2}=0,2\left(mol\right)\Rightarrow m_{CaCl_2}=0,2.111=22,2\left(g\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=0,5.1=0,5\left(mol\right)\end{matrix}\right.\)
PTHH: CaO + 2HCl ---> CaCl2 + H2O
Ban đầu: 0,2 0,5
Pư: 0,2 0,4
Sau pư: 0 0,1 0,2
=> mmuối = 0,2.111 = 22,2 (g)