Câu 5:
a, PT: \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
b, Ta có: \(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaO}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)