Ta có: \(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PT: \(RO+H_2O\rightarrow R\left(OH\right)_2\)
\(R\left(OH\right)_2+2HCl\rightarrow RCl_2+2H_2O\)
Theo PT: \(n_{RO}=n_{R\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow M_{RO}=\dfrac{11,2}{0,2}=56\left(g/mol\right)\)
⇒ MR + 16 = 56 ⇒ MX = 40 (g/mol)
→ X là Ca.
→ Đáp án: C.