nCaO=0,1(mol)
CaO + 2HCl -> CaCl2 + H2O (1)
0,1 0,2 0,1
CM dd HCl=\(\dfrac{0,2}{0,2}=1M\)
mCaCl2=111.0,1=11,1(g)
PTHH: CaO + 2HCl → CaCl2 + H2O
Đổi: 200 ml = 0,2 l
a) \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{CaO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
b) Theo PT: \(n_{CaCl_2}=n_{CaO}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCl_2}=0,1\times111=11,1\left(g\right)\)